Chapter 2 Relations and Functions (Additional Questions)

Solution:

Given the equality of two ordered pairs:

$(2a+1, 3b-2) = (5, 4)$

For two ordered pairs to be equal, their corresponding components must be equal.

Equating the first components:

$2a+1 = 5$

Solving for $a$:

$2a = 5 - 1$

$2a = 4$

$a = \frac{4}{2}$

$a = 2$

Equating the second components:

$3b-2 = 4$

Solving for $b$:

$3b = 4 + 2$

$3b = 6$

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

$b = 2$

Thus, the values of $a$ and $b$ are $a=2$ and $b=2$.

This matches option (A).

The final answer is $\boxed{a=2, b=2}$.

Therefore, the correct option is (A) $a=2, b=2$.

Solution:

Given:

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

Set $B = \{3, 4\}$

To Find:

The Cartesian product $A \times B$.

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

To find $A \times B$, we take each element from set $A$ and pair it with each element from set $B$ to form an ordered pair.

For the element $1 \in A$, the pairs are $(1, 3)$ and $(1, 4)$.

For the element $2 \in A$, the pairs are $(2, 3)$ and $(2, 4)$.

Combining all these pairs, we get the set $A \times B$:

$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$

Comparing this result with the given options, we find that it matches option (B).

The final answer is $\boxed{\{(1,3), (1,4), (2,3), (2,4)\}}$.

Therefore, the correct option is (B) $\{(1,3), (1,4), (2,3), (2,4)\}$.

Solution:

Given:

The number of elements in set A, $n(A) = 3$.

The number of elements in set B, $n(B) = 2$.

To Find:

The number of elements in the Cartesian product $A \times A \times B$, denoted as $n(A \times A \times B)$.

The number of elements in the Cartesian product of multiple finite sets is the product of the number of elements in each set.

For sets $S_1, S_2, ..., S_k$, the number of elements in their Cartesian product is given by:

$n(S_1 \times S_2 \times ... \times S_k) = n(S_1) \times n(S_2) \times ... \times n(S_k)$.

In this case, we need to find $n(A \times A \times B)$. Using the formula:

$n(A \times A \times B) = n(A) \times n(A) \times n(B)$

Substitute the given values of $n(A)$ and $n(B)$ into the formula:

$n(A \times A \times B) = 3 \times 3 \times 2$

$n(A \times A \times B) = 9 \times 2$

$n(A \times A \times B) = 18$

Thus, the number of elements in $A \times A \times B$ is 18.

The final answer is $\boxed{18}$.

Comparing this result with the given options, we find that it matches option (A).

Therefore, the correct option is (A) 18.

Solution:

Given:

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

Relation $R$ on $A$ is defined as $R = \{(x, y) : x, y \in A, y = x+1\}$.

To Find:

The range of the relation $R$.

The relation $R$ consists of ordered pairs $(x, y)$ such that both $x$ and $y$ are elements of set $A$, and $y$ is one more than $x$. We need to find the elements of $R$ by checking which pairs $(x, y)$ satisfy these conditions.

Let's list the possible ordered pairs $(x, y)$ where $x \in A$, $y = x+1$, and we also require $y \in A$.

If $x = 1$, then $y = 1+1 = 2$. Since $1 \in A$ and $2 \in A$, the pair $(1, 2)$ is in $R$.

If $x = 2$, then $y = 2+1 = 3$. Since $2 \in A$ and $3 \in A$, the pair $(2, 3)$ is in $R$.

If $x = 3$, then $y = 3+1 = 4$. Since $3 \in A$ and $4 \in A$, the pair $(3, 4)$ is in $R$.

If $x = 4$, then $y = 4+1 = 5$. Since $4 \in A$ and $5 \in A$, the pair $(4, 5)$ is in $R$.

If $x = 5$, then $y = 5+1 = 6$. Since $6 \notin A$, the pair $(5, 6)$ is not in $R$.

So, the relation $R$ is the set of ordered pairs:

$R = \{(1, 2), (2, 3), (3, 4), (4, 5)\}$

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

From the ordered pairs in $R$, the second components are $2, 3, 4, 5$.

Therefore, the range of $R$ is $\{2, 3, 4, 5\}$.

Comparing this result with the given options, we find that it matches option (B).

The final answer is $\boxed{\{2, 3, 4, 5\}}$.

Therefore, the correct option is (B) $\{2, 3, 4, 5\}$.

Solution:

Given:

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

Set $B = \{a, b, c, d\}$.

Four relations $R_1, R_2, R_3, R_4$.

To Determine:

Which of the given relations is a function from set $A$ to set $B$.

A relation $f$ from a set $A$ to a set $B$ is called a function if every element of set $A$ has one and only one image in set $B$. This means two conditions must be satisfied:

1. Every element in set $A$ must be the first element of at least one ordered pair in the relation.

2. No element in set $A$ can be the first element of more than one ordered pair in the relation (i.e., no element in $A$ can be mapped to more than one element in $B$).

Let's examine each relation:

(A) $R_1 = \{(1, a), (2, b), (3, c), (1, d)\}$

The elements of set $A$ are $1, 2, 3$.

In $R_1$, the first elements of the ordered pairs are $1, 2, 3, 1$.

The element $1 \in A$ appears as the first element in two ordered pairs: $(1, a)$ and $(1, d)$. This means that the element $1$ in set $A$ is associated with two different elements ($a$ and $d$) in set $B$. This violates the second condition for a function.

Therefore, $R_1$ is not a function from $A$ to $B$.

(B) $R_2 = \{(1, a), (2, b)\}$

The elements of set $A$ are $1, 2, 3$.

In $R_2$, the first elements of the ordered pairs are $1, 2$.

The element $3 \in A$ does not appear as the first element of any ordered pair in $R_2$. This means that the element $3$ in set $A$ does not have an image in set $B$ under the relation $R_2$. This violates the first condition for a function.

Therefore, $R_2$ is not a function from $A$ to $B$.

(C) $R_3 = \{(1, a), (2, a), (3, a)\}$

The elements of set $A$ are $1, 2, 3$.

In $R_3$, the first elements of the ordered pairs are $1, 2, 3$. Every element of set $A$ is the first element of exactly one ordered pair.

• Element $1 \in A$ is mapped to $a \in B$. (One image)
• Element $2 \in A$ is mapped to $a \in B$. (One image)
• Element $3 \in A$ is mapped to $a \in B$. (One image)

Both conditions for a function are satisfied: every element in $A$ has an image, and each element has exactly one image (even though different elements in $A$ can map to the same element in $B$).

Therefore, $R_3$ is a function from $A$ to $B$.

(D) $R_4 = \{(1, a), (2, b), (3, c), (4, d)\}$

The elements of set $A$ are $1, 2, 3$.

In $R_4$, the first elements of the ordered pairs are $1, 2, 3, 4$.

The element $4$ is the first element of the pair $(4, d)$, but $4$ is not an element of set $A = \{1, 2, 3\}$. Thus, $R_4$ is not a relation from $A$ to $B$. For a relation to be a function from $A$ to $B$, its domain must be exactly $A$. The domain of $R_4$ is $\{1, 2, 3, 4\}$, which is not equal to $A$.

Therefore, $R_4$ is not a function from $A$ to $B$.

Based on the analysis, only relation $R_3$ satisfies the conditions to be a function from $A$ to $B$.

The final answer is $\boxed{R_3 = \{(1, a), (2, a), (3, a)\}}$.

Therefore, the correct option is (C) $R_3 = \{(1, a), (2, a), (3, a)\}$.

Solution:

Given:

The real function $f(x) = \frac{1}{x^2 - 4}$.

To Find:

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

For a real function defined as a fraction, the denominator cannot be equal to zero, because division by zero is undefined in the set of real numbers.

Therefore, for $f(x) = \frac{1}{x^2 - 4}$ to be defined, the denominator $x^2 - 4$ must not be equal to zero.

We need to find the values of $x$ for which the denominator is zero:

$x^2 - 4 = 0$

This is a difference of squares, which can be factored as $(x-2)(x+2)$.

So, $(x-2)(x+2) = 0$

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

Either $x-2 = 0$ or $x+2 = 0$.

Solving for $x$ in the first equation:

$x = 2$

Solving for $x$ in the second equation:

$x = -2$

Thus, the denominator $x^2 - 4$ is equal to zero when $x = 2$ or $x = -2$. These are the values of $x$ for which the function is undefined.

The domain of the function is the set of all real numbers except the values that make the denominator zero.

Domain $= \mathbb{R} - \{-2, 2\}$

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

The final answer is $\boxed{\mathbb{R} - \{-2, 2\}}$.

Therefore, the correct option is (D) $\mathbb{R} - \{-2, 2\}$.

Solution:

Given:

The function $f(x) = |x - 1|$.

To Find:

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

The function is defined as $f(x) = |x - 1|$. The absolute value of any real number is always non-negative. That is, for any real number $y$, $|y| \geq 0$.

In this function, the expression inside the absolute value is $(x - 1)$. As $x$ takes any real value, $(x - 1)$ can take any real value (positive, negative, or zero).

However, the absolute value of $(x - 1)$, which is $|x - 1|$, will always be greater than or equal to zero.

$|x - 1| \geq 0$ for all real values of $x$.

The minimum value of $|x - 1|$ is $0$, which occurs when $x - 1 = 0$, i.e., when $x = 1$.

$f(1) = |1 - 1| = |0| = 0$.

For any other real value of $x$, $|x - 1|$ will be a positive real number. As $x$ moves away from $1$ (in either direction), $|x - 1|$ increases and can take any positive real value.

Therefore, the set of all possible values that $f(x)$ can take is the set of all non-negative real numbers.

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

Comparing this result with the given options, we find that it matches option (C).

The final answer is $\boxed{[0, \infty)}$.

Therefore, the correct option is (C) $[0, \infty)$.

Solution:

Given:

Assertion (A): The relation $R = \{(x, x^2) : x \text{ is a prime number less than } 10\}$ is a function.

Reason (R): Every element in the domain of $R$ has a unique image.

Let's first determine the relation $R$.

The prime numbers less than 10 are 2, 3, 5, and 7.

So, the set of $x$ values in the relation is $\{2, 3, 5, 7\}$. This set is the domain of the relation $R$.

Now, let's find the ordered pairs $(x, x^2)$ for each $x$ in the domain:

• For $x=2$, $x^2 = 2^2 = 4$. The ordered pair is $(2, 4)$.
• For $x=3$, $x^2 = 3^2 = 9$. The ordered pair is $(3, 9)$.
• For $x=5$, $x^2 = 5^2 = 25$. The ordered pair is $(5, 25)$.
• For $x=7$, $x^2 = 7^2 = 49$. The ordered pair is $(7, 49)$.

The relation $R$ is therefore $R = \{(2, 4), (3, 9), (5, 25), (7, 49)\}$.

Checking Assertion (A):

A relation is a function if and only if every element in its domain has exactly one image.

The domain of $R$ is $\{2, 3, 5, 7\}$.

Let's check if each element in the domain has a unique image:

• The element 2 is related only to 4. Its image is unique.
• The element 3 is related only to 9. Its image is unique.
• The element 5 is related only to 25. Its image is unique.
• The element 7 is related only to 49. Its image is unique.

Every element in the domain $\{2, 3, 5, 7\}$ appears exactly once as the first component in the ordered pairs of $R$. This means that every element in the domain has one and only one image.

Therefore, the relation $R$ is a function.

Assertion (A) is True.

Checking Reason (R):

Reason (R) states: Every element in the domain of $R$ has a unique image.

This statement is precisely the definition of a function from its domain to its range.

From our analysis above, we found that every element in the domain of $R$ (which is $\{2, 3, 5, 7\}$) does indeed have a unique image in the range (which is $\{4, 9, 25, 49\}$).

Therefore, Reason (R) is True.

Relation between A and R:

Assertion (A) claims that $R$ is a function. Reason (R) provides the condition that defines a function: every element in the domain has a unique image.

Our process for checking if $R$ is a function directly used the condition stated in Reason (R). We verified that every element in the domain of $R$ has a unique image, and based on this, we concluded that $R$ is a function.

Therefore, Reason (R) is the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The final answer is $\boxed{\text{Both A and R are true and R is the correct explanation of A}}$.

Therefore, the correct option is (A) Both A and R are true and R is the correct explanation of A.

Solution:

Given:

The function $f(x) = x^2$.

The function $g(x) = 2x+1$.

To Find:

The sum of the functions $(f+g)(x)$.

The sum of two functions $f$ and $g$, denoted by $(f+g)$, is defined for all $x$ in the intersection of the domains of $f$ and $g$ as:

$(f+g)(x) = f(x) + g(x)$

Substitute the given expressions for $f(x)$ and $g(x)$ into the definition:

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

Simplify the expression by removing the parentheses:

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

Thus, the sum of the functions $f(x)$ and $g(x)$ is $x^2 + 2x + 1$.

The final answer is $\boxed{x^2 + 2x + 1}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) $x^2 + 2x + 1$.

Solution:

We need to match each given function with its correct type.

(i) The function $f(x) = 5$. This function assigns the same constant value (5) to every input $x$. This is the definition of a Constant Function.

So, (i) matches with (b).

(ii) The function $f(x) = x$. This function assigns the input value itself as the output value. This is the definition of an Identity Function.

So, (ii) matches with (a).

(iii) The function $f(x) = |x|$. This function assigns the absolute value of the input $x$ as the output value. This is the definition of a Modulus Function.

So, (iii) matches with (c).

(iv) The function $f(x) = \lfloor x \rfloor$. This function assigns the greatest integer less than or equal to the input $x$ as the output value. This is the definition of the Greatest Integer Function.

So, (iv) matches with (d).

Based on the matching:

• (i) - (b)
• (ii) - (a)
• (iii) - (c)
• (iv) - (d)

Now let's check the given options:

(A) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d) - Incorrect

(B) (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d) - Correct

(C) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c) - Incorrect

(D) (i)-(a), (ii)-(b), (iii)-(d), (iv)-(c) - Incorrect

The final answer is $\boxed{(i)-(b), (ii)-(a), (iii)-(c), (iv)-(d)}$.

Therefore, the correct option is (B) (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d).

Solution:

Given:

The real function $f(x) = \sqrt{9 - x^2}$.

To Find:

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

For the function $f(x) = \sqrt{9 - x^2}$ to yield a real number, the expression under the square root must be non-negative (greater than or equal to zero).

So, we must have:

$9 - x^2 \geq 0$

We can rewrite the inequality as:

$-x^2 \geq -9$

Multiplying both sides by -1 and reversing the inequality sign:

$x^2 \leq 9$

Taking the square root of both sides gives:

$\sqrt{x^2} \leq \sqrt{9}$

$|x| \leq 3$

The inequality $|x| \leq k$ for $k > 0$ is equivalent to $-k \leq x \leq k$.

In this case, $k=3$. So, the inequality $|x| \leq 3$ is equivalent to:

$-3 \leq x \leq 3$

This means that $x$ must be a real number between -3 and 3, including -3 and 3.

In interval notation, the set of all such $x$ is $[-3, 3]$.

Thus, the domain of the function $f(x) = \sqrt{9 - x^2}$ is the closed interval $[-3, 3]$.

The final answer is $\boxed{[-3, 3]}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) $[-3, 3]$.

Solution:

Given:

The function $f(x) = x+1$.

The function $g(x) = x-1$.

To Find:

The product of the functions $(f \cdot g)(x)$.

The product of two functions $f$ and $g$, denoted by $(f \cdot g)$, is defined for all $x$ in the intersection of the domains of $f$ and $g$ as:

$(f \cdot g)(x) = f(x) \cdot g(x)$

Substitute the given expressions for $f(x)$ and $g(x)$ into the definition:

$(f \cdot g)(x) = (x+1) \cdot (x-1)$

Use the difference of squares formula, which states that $(a+b)(a-b) = a^2 - b^2$. In this case, $a=x$ and $b=1$.

$(f \cdot g)(x) = x^2 - 1^2$

$(f \cdot g)(x) = x^2 - 1$

Thus, the product of the functions $f(x)$ and $g(x)$ is $x^2 - 1$.

The final answer is $\boxed{x^2 - 1}$.

Comparing this result with the given options, we find that it matches option (A).

Therefore, the correct option is (A) $x^2 - 1$.

Solution:

We are asked to identify the statement that is NOT true among the given options regarding operations on functions $f$ and $g$. Let $D_f$ be the domain of $f$ and $D_g$ be the domain of $g$.

Let's review the definitions of the basic operations on functions:

(A) Sum of functions: The sum of $f$ and $g$, denoted by $f+g$, is defined by:

$(f+g)(x) = f(x) + g(x)$

This definition is valid for all $x$ in the intersection of the domains of $f$ and $g$, i.e., for $x \in D_f \cap D_g$.

Statement (A) is the standard definition and is True.

(B) Difference of functions: The difference of $f$ and $g$, denoted by $f-g$, is defined by:

$(f-g)(x) = f(x) - g(x)$

This definition is valid for all $x$ in the intersection of the domains of $f$ and $g$, i.e., for $x \in D_f \cap D_g$.

Statement (B) is the standard definition and is True.

(C) Product of functions: The product of $f$ and $g$, denoted by $f \cdot g$, is defined by:

$(f \cdot g)(x) = f(x) \cdot g(x)$

This definition is valid for all $x$ in the intersection of the domains of $f$ and $g$, i.e., for $x \in D_f \cap D_g$.

Statement (C) is the standard definition and is True.

(D) Quotient of functions: The quotient of $f$ and $g$, denoted by $\frac{f}{g}$, is defined by:

$(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$

This definition is valid for all $x$ in the intersection of the domains of $f$ and $g$, such that $g(x) \neq 0$. The domain of $(\frac{f}{g})(x)$ is $\{x \in D_f \cap D_g \mid g(x) \neq 0\}$.

Statement (D) says $(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$, where $g(x) = 0$. This statement implies that the quotient function is defined even when the denominator $g(x)$ is zero. However, division by zero is undefined in mathematics.

Therefore, the equation $(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$ is not defined for values of $x$ where $g(x) = 0$. The statement claiming it holds "where $g(x) = 0$" is fundamentally incorrect because the function itself is not defined at such points.

Statement (D) is False.

Based on the analysis, the statement that is NOT true is (D).

The final answer is $\boxed{(\frac{f}{g})(x) = \frac{f(x)}{g(x)}, \text{ where } g(x) = 0}$.

Therefore, the correct option is (D) $(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$, where $g(x) = 0$.

Solution:

Given:

The Signum function, $f(x) = \begin{cases} 1 & , & x > 0 \\ 0 & , & x = 0 \\ -1 & , & x < 0 \end{cases}$.

To Find:

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

The range of a function is the set of all possible output values ($f(x)$) that the function can produce for inputs in its domain.

The domain of the Signum function is all real numbers, $\mathbb{R}$.

We examine the definition of the function for different intervals of $x$:

1. When $x > 0$ (e.g., $x=5$, $x=0.1$), the function definition states that $f(x) = 1$. So, 1 is a possible output value.

2. When $x = 0$, the function definition states that $f(x) = 0$. So, 0 is a possible output value.

3. When $x < 0$ (e.g., $x=-2$, $x=-0.001$), the function definition states that $f(x) = -1$. So, -1 is a possible output value.

These are the only three cases for any real number $x$. Therefore, the only possible output values of the function $f(x)$ are 1, 0, and -1.

The set of all possible output values is $\{-1, 0, 1\}$.

The range of the Signum function is the set $\{-1, 0, 1\}$.

The final answer is $\boxed{\{1, 0, -1\}}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) $\{1, 0, -1\}$.

Solution:

Given:

Set of students $A = \{\text{Amit, Bina, Chirag, Divya, Eshan}\}$.

Relation $R$ from $A$ to $B$ defined by $(s, m) \in R$ if student $s$ obtained $m$ marks.

The given data is: Amit (45), Bina (38), Chirag (45), Divya (40), Eshan (38).

To Determine:

Whether the relation $R$ is a function and identify the correct statement about it.

The relation $R$ is a set of ordered pairs $(s, m)$, where $s$ is a student from set $A$ and $m$ is the mark obtained. Based on the given data, we can write the relation $R$ as:

$R = \{(\text{Amit}, 45), (\text{Bina}, 38), (\text{Chirag}, 45), (\text{Divya}, 40), (\text{Eshan}, 38)\}$

For a relation from set $A$ to set $B$ to be a function, two conditions must be satisfied:

1. Every element in the domain (set $A$) must be associated with some element in the codomain (set $B$). In terms of ordered pairs, every element of $A$ must appear as the first component in at least one ordered pair in the relation.

2. Every element in the domain (set $A$) must be associated with a unique element in the codomain (set $B$). In terms of ordered pairs, no element of $A$ can appear as the first component in more than one ordered pair.

Let's check these conditions for the given relation $R$ with domain $A = \{\text{Amit, Bina, Chirag, Divya, Eshan}\}$.

1. Check if every student in $A$ is in the domain of $R$ (appears as a first component):

• Amit is the first component in $(\text{Amit}, 45)$.
• Bina is the first component in $(\text{Bina}, 38)$.
• Chirag is the first component in $(\text{Chirag}, 45)$.
• Divya is the first component in $(\text{Divya}, 40)$.
• Eshan is the first component in $(\text{Eshan}, 38)$.

All elements in set $A$ appear as the first component. So, condition 1 is satisfied.

2. Check if each student in $A$ is associated with a unique mark (appears as a first component in only one ordered pair):

• Amit appears only in $(\text{Amit}, 45)$. Unique image (45).
• Bina appears only in $(\text{Bina}, 38)$. Unique image (38).
• Chirag appears only in $(\text{Chirag}, 45)$. Unique image (45).
• Divya appears only in $(\text{Divya}, 40)$. Unique image (40).
• Eshan appears only in $(\text{Eshan}, 38)$. Unique image (38).

Each student in set $A$ appears exactly once as the first component. This means each student has one and only one mark associated with them according to the relation $R$. So, condition 2 is satisfied.

Since both conditions are satisfied, the relation $R$ is a function from $A$ to $B$.

It is important to note that having different students getting the same mark (e.g., Amit and Chirag both got 45) does not violate the definition of a function. This is a characteristic of a many-to-one function, which is a valid type of function.

Now let's evaluate the given statements:

(A) It is a function because each student is mapped to a unique mark.

This statement correctly identifies the relation as a function and provides the correct reason based on the definition of a function. Every element in the domain (student) is mapped to exactly one element in the codomain (mark).

(B) It is a function because each mark is associated with a unique student.

This statement describes a one-to-one function. The relation $R$ is not one-to-one because the marks 45 and 38 are associated with more than one student. Thus, this statement is incorrect.

(C) It is a relation but not a function because Chirag and Amit have the same mark.

As explained earlier, having the same mark for different students does not prevent a relation from being a function. This statement is incorrect.

(D) It is a relation but not a function because two students have the same mark.

Similar to (C), this statement provides an incorrect reason for the relation not being a function. This statement is incorrect.

Based on our analysis, statement (A) is the true statement.

The final answer is $\boxed{\text{It is a function because each student is mapped to a unique mark}}$.

Therefore, the correct option is (A) It is a function because each student is mapped to a unique mark.

Solution:

Given:

The relation $R$ is from the set of students $A$ to the set of marks $B$, defined by $(s, m) \in R$ if student $s$ obtained $m$ marks.

The relation is represented by the following data for 5 students:

To Find:

The domain of the relation $R$.

The relation $R$ can be written as a set of ordered pairs $(s, m)$:

$R = \{(\text{Amit}, 45), (\text{Bina}, 38), (\text{Chirag}, 45), (\text{Divya}, 40), (\text{Eshan}, 38)\}$

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

In the ordered pairs of $R$, the first components are the names of the students: Amit, Bina, Chirag, Divya, Eshan.

Therefore, the domain of the relation $R$ is the set containing these names.

Domain of $R = \{\text{Amit, Bina, Chirag, Divya, Eshan}\}$

Comparing this set with the given options:

(A) $\{$Amit, Bina, Chirag, Divya, Eshan$\}$ - This matches our calculated domain.

(B) $\{38, 40, 45\}$ - This is the set of all second components, which is the range of the relation.

(C) $\{38, 40, 45\}$, $\{$Amit, Bina, Chirag, Divya, Eshan$\}$ - This lists both the range and the domain, not just the domain.

(D) The set of all students in the class. The problem states that $A$ is the set of students in the class, and based on the data, the domain of $R$ is equal to $A$. Option (A) lists the elements of $A$. Both (A) and (D) represent the same set, but (A) explicitly lists the elements derived from the relation data.

Option (A) is the explicit set of elements that constitute the domain of the relation $R$ as given by the data.

The final answer is $\boxed{\{\text{Amit, Bina, Chirag, Divya, Eshan}\}}$.

Therefore, the correct option is (A) $\{$Amit, Bina, Chirag, Divya, Eshan$\}$.

Solution:

Given:

The relation $R$ from the set of students $A$ to the set of marks $B$, defined by $(s, m) \in R$ if student $s$ obtained $m$ marks.

The relation is represented by the following data for 5 students:

To Find:

The range of the relation $R$.

The relation $R$ can be written as a set of ordered pairs $(s, m)$, where $s$ is a student and $m$ is their mark:

$R = \{(\text{Amit}, 45), (\text{Bina}, 38), (\text{Chirag}, 45), (\text{Divya}, 40), (\text{Eshan}, 38)\}$

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

In the ordered pairs of $R$, the second components (the marks obtained) are: 45, 38, 45, 40, 38.

To find the range, we collect all the unique second components into a set.

The unique second components are 38, 40, and 45.

Therefore, the range of the relation $R$ is the set $\{38, 40, 45\}$.

Comparing this set with the given options:

(A) $\{$Amit, Bina, Chirag, Divya, Eshan$\}$ - This is the domain of the relation.

(B) $\{38, 40, 45\}$ - This matches our calculated range.

(C) $\{38, 40, 45\}$, $\{$Amit, Bina, Chirag, Divya, Eshan$\}$ - This lists both the range and the domain.

(D) The set of all possible marks out of 50. This is the codomain, which could be $\{0, 1, 2, ..., 50\}$. The range is a subset of the codomain containing only the marks actually obtained by the students in the relation.

Based on our calculation, the range of the relation $R$ is $\{38, 40, 45\}$.

The final answer is $\boxed{\{38, 40, 45\}}$.

Therefore, the correct option is (B) $\{38, 40, 45\}$.

Solution:

Given:

Set $A = \{p, q, r\}$.

Set $B = \{1, 2\}$.

To Determine:

Which of the given options is a relation from set $A$ to set $B$.

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

First, let's find the Cartesian product $A \times B$. This is the set of all ordered pairs $(a, b)$ where the first element $a$ is from set $A$, and the second element $b$ is from set $B$.

$A \times B = \{(x, y) \mid x \in A \text{ and } y \in B\}$

Using the given sets $A = \{p, q, r\}$ and $B = \{1, 2\}$:

$A \times B = \{(p, 1), (p, 2), (q, 1), (q, 2), (r, 1), (r, 2)\}$

Now, we check each given option to see if it is a subset of $A \times B$. A set is a subset of another if all its elements are contained within the other set.

(A) $\{(p, 1), (q, 2), (r, 1), (p, 2)\}$

Let's check each ordered pair in this set:

• $(p, 1)$: $p \in A$ and $1 \in B$. So, $(p, 1) \in A \times B$.
• $(q, 2)$: $q \in A$ and $2 \in B$. So, $(q, 2) \in A \times B$.
• $(r, 1)$: $r \in A$ and $1 \in B$. So, $(r, 1) \in A \times B$.
• $(p, 2)$: $p \in A$ and $2 \in B$. So, $(p, 2) \in A \times B$.

Since all the ordered pairs in option (A) are elements of $A \times B$, the set in option (A) is a subset of $A \times B$.

Therefore, option (A) is a relation from $A$ to $B$.

(B) $\{(1, p), (2, q)\}$

Let's check the first ordered pair:

• $(1, p)$: $1 \in B$ but $1 \notin A$. The first element must be from $A$ for the pair to be in $A \times B$. So, $(1, p) \notin A \times B$.

Since not all elements are in $A \times B$, option (B) is not a relation from $A$ to $B$. (Note: This is a relation from $B$ to $A$).

(C) $\{(p, a), (q, b)\}$ where $a, b \in B$

This option describes a set of ordered pairs where the first components are $p$ and $q$ (which are in $A$), and the second components $a$ and $b$ are from $B = \{1, 2\}$. Any pair $(p, a)$ with $a \in B$ is in $A \times B$, and any pair $(q, b)$ with $b \in B$ is in $A \times B$. Thus, a set of the form $\{(p, a), (q, b)\}$ where $a, b \in B$ is indeed a subset of $A \times B$. Examples include $\{(p, 1), (q, 1)\}$, $\{(p, 1), (q, 2)\}$, $\{(p, 2), (q, 1)\}$, $\{(p, 2), (q, 2)\}$. While this describes a form of relation from A to B, option (A) provides a specific, concrete set that is a relation.

(D) $\{(1, p), (2, q), (3, r)\}$

Let's check the ordered pairs:

• $(1, p)$: $1 \notin A$. So, $(1, p) \notin A \times B$.
• $(2, q)$: $2 \notin A$. So, $(2, q) \notin A \times B$.
• $(3, r)$: $3$ is not an element of $A$ or $B$. So, $(3, r) \notin A \times B$.

Since none of the elements are in $A \times B$, option (D) is not a relation from $A$ to $B$.

Comparing the options, only option (A) provides a set where all ordered pairs have the first component from $A$ and the second component from $B$. Thus, only (A) is a subset of $A \times B$ and is a relation from $A$ to $B$.

The final answer is $\boxed{\{(p, 1), (q, 2), (r, 1), (p, 2)\}}$.

Therefore, the correct option is (A) $\{(p, 1), (q, 2), (r, 1), (p, 2)\}$.

Solution:

Given:

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

To Find:

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

The domain of a real function is the set of all real numbers for which the function is defined.

For a rational function (a function expressed as a ratio of two polynomials), the function is defined everywhere except where the denominator is equal to zero, because division by zero is undefined.

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

The function is undefined when the denominator is zero.

Set the denominator equal to zero to find the values of $x$ that are excluded from the domain:

$x - 1 = 0$

Solving for $x$:

$x = 1$

This means that the function $f(x)$ is undefined when $x=1$. For all other real values of $x$, the denominator $x-1$ is non-zero, and the function is defined.

Therefore, the domain of the function $f(x)$ is the set of all real numbers ($\mathbb{R}$) excluding the value $1$.

Domain of $f = \mathbb{R} - \{1\}$

Comparing this result with the given options, we find that it matches option (A).

The final answer is $\boxed{\mathbb{R} - \{1\}}$.

Therefore, the correct option is (A) $\mathbb{R} - \{1\}$.

Solution:

To Determine:

Which of the given graphs represents a function.

A graph represents a function if and only if every vertical line intersects the graph at most once. This is known as the Vertical Line Test.

Let's analyze each option based on the Vertical Line Test:

(A) Graph A (Vertical line passing through two points on the curve)

The description states that a vertical line passes through two points on the curve. This means there exists at least one vertical line that intersects the graph at more than one point. According to the Vertical Line Test, this graph does not represent a function.

(B) Graph B (Circle)

For a circle that is not a single point, it is possible to draw a vertical line that intersects the circle at two distinct points (an upper half and a lower half for the same x-coordinate, within the diameter). For example, the graph of $x^2 + y^2 = r^2$ (a circle centered at the origin with radius $r$) fails the vertical line test for any $x$ such that $-r < x < r$, where $y = \pm\sqrt{r^2 - x^2}$. Thus, a circle does not represent a function of $x$.

(C) Graph C (Parabola opening upwards)

A parabola opening upwards, like the graph of $y = x^2$, has the property that for every $x$ value in its domain, there is exactly one corresponding $y$ value. Graphically, any vertical line drawn across a parabola opening upwards will intersect the parabola at exactly one point. Therefore, a parabola opening upwards represents a function.

(D) Graph D (Parabola opening sideways)

A parabola opening sideways, like the graph of $x = y^2$, has the property that for a given $x$ value (specifically, for $x > 0$ in the case of $x=y^2$), there can be two corresponding $y$ values (a positive and a negative value, e.g., for $x=4$, $y=\pm 2$). Graphically, a vertical line can intersect a parabola opening sideways at two distinct points. Therefore, a parabola opening sideways does not represent a function of $x$.

Based on the Vertical Line Test, only the graph of a parabola opening upwards represents a function among the described options.

The final answer is $\boxed{\text{Graph C (Parabola opening upwards)}}$.

Therefore, the correct option is (C) Graph C (Parabola opening upwards).

Solution:

Given:

Set $A$ with $n(A) = 3$ elements.

Set $B$ with $n(B) = 2$ elements.

To Find:

The number of relations from set $A$ to set $B$.

A relation from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$.

First, we find the number of elements in the Cartesian product $A \times B$. If set $A$ has $m$ elements and set $B$ has $n$ elements, then the number of elements in $A \times B$ is $m \times n$.

$n(A \times B) = n(A) \times n(B)$

Substitute the given values $n(A)=3$ and $n(B)=2$:

$n(A \times B) = 3 \times 2 = 6$

Now, the number of relations from $A$ to $B$ is equal to the number of possible subsets of $A \times B$. The number of subsets of a set with $k$ elements is $2^k$.

Number of relations from $A$ to $B = \text{Number of subsets of } A \times B = 2^{n(A \times B)}$

Substitute the calculated value of $n(A \times B)$:

Number of relations = $2^6$

Comparing this result with the given options:

(A) $2^{3+2} = 2^5$

(B) $2^{3 \times 2} = 2^6$

(C) $3^2 = 9$

(D) $2^3 = 8$

The calculated number of relations, $2^6$, matches option (B) because $3 \times 2 = 6$.

The final answer is $\boxed{2^{3 \times 2}}$.

Therefore, the correct option is (B) $2^{3 \times 2}$.

Solution:

Given:

Function $f(x) = x^2$.

Function $g(x) = \sqrt{x}$.

To Find:

The domain of the function $(\frac{f}{g})(x)$.

The quotient of two functions $f$ and $g$, denoted by $(\frac{f}{g})$, is defined by:

$(\frac{f}{g})(x) = \frac{f(x)}{g(x)}$

The domain of $(\frac{f}{g})(x)$ is the set of all $x$ that are in the domain of $f$ AND in the domain of $g$, with the additional condition that $g(x) \neq 0$.

Domain$(\frac{f}{g}) = \{x \mid x \in \text{Domain}(f) \text{ and } x \in \text{Domain}(g) \text{ and } g(x) \neq 0\}$

First, find the domain of $f(x) = x^2$.

Since $f(x)$ is a polynomial, it is defined for all real numbers.

Domain$(f) = \mathbb{R}$ or $(-\infty, \infty)$.

Next, find the domain of $g(x) = \sqrt{x}$.

For $\sqrt{x}$ to be a real number, the expression under the square root must be non-negative.

So, $x \geq 0$.

Domain$(g) = \{x \in \mathbb{R} \mid x \geq 0\}$ or $[0, \infty)$.

Now, consider the condition $g(x) \neq 0$.

$g(x) = \sqrt{x}$

We need $\sqrt{x} \neq 0$.

$\sqrt{x} = 0$ only when $x = 0$.

So, we must exclude $x=0$ from the domain.

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

Domain$(\frac{f}{g}) = \text{Domain}(f) \cap \text{Domain}(g) \cap \{x \mid g(x) \neq 0\}$

Domain$(\frac{f}{g}) = (-\infty, \infty) \cap [0, \infty) \cap \{x \mid x \neq 0\}$

The intersection of $(-\infty, \infty)$ and $[0, \infty)$ is $[0, \infty)$.

Now, we exclude $x=0$ from $[0, \infty)$.

$[0, \infty) - \{0\} = (0, \infty)$

Thus, the domain of $(\frac{f}{g})(x)$ is $(0, \infty)$.

The final answer is $\boxed{(0, \infty)}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) $(0, \infty)$.

Solution:

Given:

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

To Find:

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

For the function $f(x)$ to be defined as a real number, two conditions must be met:

1. The expression under the square root must be non-negative.

2. The denominator cannot be equal to zero.

From condition 1, the expression inside the square root, $x-5$, must be greater than or equal to zero:

$x - 5 \geq 0$

Solving for $x$:

$x \geq 5$

From condition 2, the denominator $\sqrt{x-5}$ must not be equal to zero.

$\sqrt{x-5} \neq 0$

This means $x-5 \neq 0$.

Solving for $x$:

$x \neq 5$

We need to satisfy both conditions simultaneously. We need $x \geq 5$ AND $x \neq 5$.

Combining these two conditions, we get $x > 5$.

In interval notation, the set of all real numbers $x$ such that $x > 5$ is $(5, \infty)$.

Thus, the domain of the function $f(x) = \frac{1}{\sqrt{x-5}}$ is $(5, \infty)$.

The final answer is $\boxed{(5, \infty)}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) $(5, \infty)$.

Solution:

Given:

The function $f(x) = 3x-2$.

To Find:

The value of $f(0)$.

To find the value of the function at $x=0$, we need to substitute $x=0$ into the expression for $f(x)$.

$f(x) = 3x - 2$

Substitute $x=0$ into the equation:

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

Perform the multiplication and subtraction:

$f(0) = 0 - 2$

$f(0) = -2$

Thus, the value of $f(0)$ is -2.

The final answer is $\boxed{-2}$.

Comparing this result with the given options, we find that it matches option (C).

Therefore, the correct option is (C) -2.

Solution:

To Determine:

Which of the given statements is true.

Let's analyze each statement:

(A) Every relation is a function.

A relation is any subset of the Cartesian product of two sets. A function is a special type of relation where every element in the domain is mapped to exactly one element in the codomain.

Consider the relation $R = \{(1, a), (1, b)\}$ from set $A = \{1\}$ to set $B = \{a, b\}$. This is a relation because it is a subset of $A \times B = \{(1, a), (1, b)\}$. However, it is not a function because the element $1 \in A$ is related to two different elements, $a$ and $b$, in $B$.

Therefore, this statement is false.

(B) Every function is a relation.

By definition, a function $f$ from a set $A$ to a set $B$ is a relation from $A$ to $B$ such that every element of $A$ is related to exactly one element of $B$. This means that the set of ordered pairs defining the function is a subset of the Cartesian product $A \times B$, which is the definition of a relation from $A$ to $B$.

Therefore, every function is indeed a relation.

This statement is true.

(C) A relation from $A$ to $B$ is a subset of $B \times A$.

A relation from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$. The ordered pairs in $A \times B$ are of the form $(a, b)$ where $a \in A$ and $b \in B$. The ordered pairs in $B \times A$ are of the form $(b, a)$ where $b \in B$ and $a \in A$.

Unless $A=B$ or $A$ or $B$ is empty, $A \times B$ and $B \times A$ are different sets. A relation from $A$ to $B$ is a subset of $A \times B$, not $B \times A$.

Therefore, this statement is false.

(D) The range of a relation is the set of all first elements of the ordered pairs.

The domain of a relation is the set of all first components (elements from the first set) of the ordered pairs in the relation.

The range of a relation is the set of all second components (elements from the second set) of the ordered pairs in the relation.

Therefore, this statement incorrectly defines the range; it describes the domain.

This statement is false.

Based on the analysis, only statement (B) is true.

The final answer is $\boxed{\text{Every function is a relation}}$.

Therefore, the correct option is (B) Every function is a relation.

Solution:

Given:

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

Relation $R = \{(1, 1), (2, 2), (3, 3)\}$ on $A$. A relation on $A$ means a relation from $A$ to $A$.

To Find:

The domain and the range of the relation $R$.

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

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

The given relation is $R = \{(1, 1), (2, 2), (3, 3)\}$.

To find the domain, we list all the unique first components from the ordered pairs in $R$:

The first components are 1, 2, and 3.

Domain of $R = \{1, 2, 3\}$

To find the range, we list all the unique second components from the ordered pairs in $R$:

The second components are 1, 2, and 3.

Range of $R = \{1, 2, 3\}$

So, the domain of $R$ is $\{1, 2, 3\}$ and the range of $R$ is $\{1, 2, 3\}$.

Note that in this specific case, the domain and range are equal to the set $A$.

Comparing our results with the given options:

(A) Domain = $\{1, 2, 3\}$, Range = $\{1, 2, 3\}$ - This matches our calculation.

(B) Domain = $\{1, 2, 3\}$, Range = $\{1, 2, 3, 4, 5, ...\}$ - The range is incorrect.

(C) Domain = $A \times A$, Range = $A \times A$ - $A \times A$ is the Cartesian product, not the domain or range.

(D) Domain = $\{1, 2, 3\}$, Range = $\{1\}$ - The range is incorrect.

The final answer is $\boxed{\text{Domain = } \{1, 2, 3\}, \text{ Range = } \{1, 2, 3\}}$.

Therefore, the correct option is (A) Domain = $\{1, 2, 3\}$, Range = $\{1, 2, 3\}$.

Solution:

Given:

The function $f(x) = x^2 + 2x - 1$.

To Find:

The value of $f(-1)$.

To find the value of the function at $x=-1$, we substitute $x=-1$ into the expression for $f(x)$.

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

Substitute $x=-1$ into the equation:

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

Evaluate the terms:

$(-1)^2 = 1$

$2(-1) = -2$

So, the expression becomes:

$f(-1) = 1 + (-2) - 1$

$f(-1) = 1 - 2 - 1$

Perform the subtraction:

$f(-1) = -1 - 1$

$f(-1) = -2$

Thus, the value of $f(-1)$ is -2.

The final answer is $\boxed{-2}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) -2.

Solution:

Given:

The identity function $f(x) = x$.

To Determine:

The characteristic of the graph of the identity function $f(x) = x$.

The graph of a function $f(x)$ is the set of all points $(x, y)$ in the Cartesian plane such that $y = f(x)$. For the identity function, $y = f(x) = x$. So the graph consists of all points $(x, y)$ where $y = x$.

Let's consider some points $(x, y)$ that satisfy $y=x$:

• If $x=0$, $y=0$. Point is $(0, 0)$. This point is the origin.
• If $x=1$, $y=1$. Point is $(1, 1)$.
• If $x=2$, $y=2$. Point is $(2, 2)$.
• If $x=-1$, $y=-1$. Point is $(-1, -1)$.

Plotting these points suggests that they lie on a straight line. The equation $y = x$ is the equation of a straight line in the form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

In the equation $y = x$, the slope $m = 1$ and the y-intercept $c = 0$.

A y-intercept of 0 means the line passes through the point where $x=0$ and $y=0$, which is the origin $(0, 0)$.

A slope of 1 means that for every 1 unit increase in $x$, the value of $y$ increases by 1 unit. The angle the line makes with the positive x-axis is $45^\circ$ since $\tan(45^\circ) = 1$.

Let's evaluate the options:

(A) Parallel to the x-axis: A line parallel to the x-axis has an equation of the form $y = c$ (where $c$ is a constant, and the slope is 0). This is not $y=x$. So, this is incorrect.

(B) Parallel to the y-axis: A line parallel to the y-axis has an equation of the form $x = c$ (where $c$ is a constant, and the slope is undefined). This is not $y=x$. So, this is incorrect.

(C) Passing through the origin with slope 1: As we found, the line $y=x$ has a y-intercept of 0, meaning it passes through the origin $(0, 0)$, and its slope is 1. This matches our analysis.

(D) Passing through the origin with slope -1: The line passing through the origin with slope -1 has the equation $y = -x$. This is not $y=x$. So, this is incorrect.

Therefore, the graph of the identity function $f(x) = x$ is a straight line passing through the origin with slope 1.

The final answer is $\boxed{\text{Passing through the origin with slope 1}}$.

Therefore, the correct option is (C) Passing through the origin with slope 1.

Solution:

Given:

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

To Find:

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

The domain of a real function is the set of all real numbers for which the function is defined as a real number.

For the function $f(x) = \frac{1}{|x|}$ to be defined, the denominator $|x|$ cannot be equal to zero, because division by zero is undefined.

So, we must have:

$|x| \neq 0$

The absolute value of a real number $x$, denoted by $|x|$, is defined as:

$|x| = x$ if $x \geq 0$

$|x| = -x$ if $x < 0$

The absolute value $|x|$ is equal to zero if and only if $x$ is equal to zero.

$|x| = 0 \iff x = 0$

Since we require $|x| \neq 0$, this means we must exclude the value of $x$ for which $|x|$ is zero.

Therefore, $x \neq 0$.

The domain of the function $f(x) = \frac{1}{|x|}$ is the set of all real numbers ($\mathbb{R}$) except for the value $0$.

Domain of $f = \mathbb{R} - \{0\}$

Comparing this result with the given options, we find that it matches option (B).

The final answer is $\boxed{\mathbb{R} - \{0\}}$.

Therefore, the correct option is (B) $\mathbb{R} - \{0\}$.

Solution:

Given:

The function $f(x) = c$, where $c$ is a constant.

To Find:

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

The range of a function is the set of all possible output values that the function can produce for inputs in its domain.

The given function is a constant function. The definition $f(x) = c$ means that for any input value of $x$ from the domain, the output value $f(x)$ is always the same constant value, $c$.

For example, if the domain is all real numbers $\mathbb{R}$:

• $f(1) = c$
• $f(0) = c$
• $f(-5) = c$
• $f(\sqrt{2}) = c$

Regardless of what real number we substitute for $x$, the function always returns the value $c$.

Therefore, the set of all possible output values of the function $f(x) = c$ is the set containing only the element $c$.

Range of $f = \{c\}$

Comparing this result with the given options:

(A) $\mathbb{R}$ - This represents all real numbers, which is the domain of a constant function, not its range (unless $c$ is defined in a way that encompasses all real numbers, which is not implied here).

(B) $\{c\}$ - This is the set containing the single constant value $c$, which is our calculated range.

(C) $[c, \infty)$ - This represents all real numbers greater than or equal to $c$. This is incorrect for a constant function.

(D) $(-\infty, c]$ - This represents all real numbers less than or equal to $c$. This is incorrect for a constant function.

The final answer is $\boxed{\{c\}}$.

Therefore, the correct option is (B) $\{c\}$.

Solution:

Given:

The relation $R$ on the set $\{1, 2, 3, 4\}$ is given by $R = \{(1, 2), (2, 3), (3, 4)\}$.

To Find:

The range of the relation $R$.

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

The given relation $R$ consists of the following ordered pairs:

• $(1, 2)$
• $(2, 3)$
• $(3, 4)$

The second components of these ordered pairs are the values $2$, $3$, and $4$.

Collecting the unique second components into a set gives the range of $R$:

Range of $R = \{2, 3, 4\}$

Comparing this result with the given options, we find that it matches option (C).

The final answer is $\boxed{\{2, 3, 4\}}$.

Therefore, the correct option is (C) $\{2, 3, 4\}$.

Solution:

Given:

The function $f(x) = 2x+3$.

The function $g(x) = x-2$.

To Find:

The value of $(f-g)(4)$.

The difference of two functions $f$ and $g$, denoted by $(f-g)$, is defined for all $x$ in the intersection of the domains of $f$ and $g$ as:

$(f-g)(x) = f(x) - g(x)$

Substitute the given expressions for $f(x)$ and $g(x)$:

$(f-g)(x) = (2x+3) - (x-2)$

Remove the parentheses and simplify the expression:

$(f-g)(x) = 2x + 3 - x + 2$

$(f-g)(x) = (2x - x) + (3 + 2)$

$(f-g)(x) = x + 5$

Now, we need to find the value of $(f-g)(4)$. Substitute $x=4$ into the expression $x+5$:

$(f-g)(4) = 4 + 5$

$(f-g)(4) = 9$

The calculated value of $(f-g)(4)$ is 9.

Let's compare this result with the given options:

(A) 13

(B) 7

(C) 5

(D) 4

The calculated result (9) is not present among the given options. There appears to be an error in the provided options for the question as stated.

However, if the question intended to ask for the value of the composite function $(f \circ g)(4)$, the calculation would be:

$(f \circ g)(x) = f(g(x)) = f(x-2) = 2(x-2) + 3 = 2x - 4 + 3 = 2x - 1$

$(f \circ g)(4) = 2(4) - 1 = 8 - 1 = 7$.

This result (7) matches option (B).

Given that a multiple-choice answer is expected, it is probable that the intended operation was composition $(f \circ g)$ instead of difference $(f-g)$. Assuming this likely typographical error in the question's notation, the intended answer would be 7.

Based on the literal reading of the question, the answer is 9, which is not in the options. Based on the options provided, the question likely intended to ask for $(f \circ g)(4)$. Assuming the latter intention, the correct option is (B).

The final answer calculated as per the question is 9.

Assuming a typo in the question and that $(f \circ g)(4)$ was intended, the result is 7.

Assuming the question meant to ask for $(f \circ g)(4)$ to align with the options, the correct option is (B) 7.

Solution:

Given:

Set of days $D = \{\text{Monday, Tuesday, Wednesday, Thursday, Friday}\}$.

Set $N$ containing the number of bottles sold on these days.

Relation $R$ from $D$ to $N$ is given by $R = \{(d, n) : \text{on day } d, n \text{ bottles were sold}\}$.

The data is:

To Determine:

Whether the relation $R$ is a function and the correct reasoning.

The relation $R$ can be written as a set of ordered pairs $(d, n)$, where $d$ is a day from set $D$ and $n$ is the number of bottles sold on that day.

$R = \{(\text{Monday}, 20), (\text{Tuesday}, 15), (\text{Wednesday}, 20), (\text{Thursday}, 25), (\text{Friday}, 30)\}$

For a relation from set $D$ to set $N$ to be a function, two conditions must be satisfied:

1. Every element in the domain (set $D$) must be associated with some element in the codomain (set $N$). Every day in $D$ must appear as the first component in at least one ordered pair in $R$.

2. Every element in the domain (set $D$) must be associated with a unique element in the codomain (set $N$). No day in $D$ can appear as the first component in more than one ordered pair.

Let's check these conditions for the given relation $R$ with domain $D = \{\text{Monday, Tuesday, Wednesday, Thursday, Friday}\}$.

1. Check if every day in $D$ is in the domain of $R$ (appears as a first component):

• Monday is the first component in $(\text{Monday}, 20)$.
• Tuesday is the first component in $(\text{Tuesday}, 15)$.
• Wednesday is the first component in $(\text{Wednesday}, 20)$.
• Thursday is the first component in $(\text{Thursday}, 25)$.
• Friday is the first component in $(\text{Friday}, 30)$.

All elements in set $D$ appear as the first component. So, condition 1 is satisfied.

2. Check if each day in $D$ is associated with a unique number of bottles sold (appears as a first component in only one ordered pair):

• Monday appears only in $(\text{Monday}, 20)$. Unique image (20).
• Tuesday appears only in $(\text{Tuesday}, 15)$. Unique image (15).
• Wednesday appears only in $(\text{Wednesday}, 20)$. Unique image (20).
• Thursday appears only in $(\text{Thursday}, 25)$. Unique image (25).
• Friday appears only in $(\text{Friday}, 30)$. Unique image (30).

Each day in set $D$ appears exactly once as the first component. This means each day is associated with one and only one number of bottles sold according to the relation $R$. So, condition 2 is satisfied.

Since both conditions are satisfied, the relation $R$ is a function from $D$ to $N$.

The fact that the same number of bottles (20) was sold on different days (Monday and Wednesday) does not violate the definition of a function. This means that different elements in the domain (Monday, Wednesday) are mapped to the same element in the codomain (20). This is characteristic of a many-to-one function, which is a valid type of function.

Now let's evaluate the given statements:

(A) Yes, because each day is mapped to a unique number of bottles sold.

This statement correctly identifies the relation as a function. The reason given, "each day is mapped to a unique number of bottles sold," is the correct reason why it is a function. 'Unique image' means each element in the domain has only one corresponding element in the codomain, which is true here.

(B) Yes, because each number of bottles sold is mapped to a unique day.

This statement describes a one-to-one function. The relation is not one-to-one because the number 20 is associated with two different days (Monday and Wednesday), and 38 is associated with two different days (Bina and Eshan in the previous problem, if we were considering that data). This statement is incorrect.

(C) No, because the same number of bottles (20) was sold on Monday and Wednesday.

This describes a situation where different elements in the domain map to the same element in the codomain, which is allowed in a function (specifically, a many-to-one function). This is not a reason for a relation to NOT be a function. This statement is incorrect.

(D) No, because the number of bottles sold is a numerical value.

The type of value in the codomain (numerical or otherwise) does not determine whether a relation is a function. This statement is incorrect.

Based on our analysis, statement (A) is the true statement.

The final answer is $\boxed{\text{Yes, because each day is mapped to a unique number of bottles sold}}$.

Therefore, the correct option is (A) Yes, because each day is mapped to a unique number of bottles sold.

Solution:

Given:

The real function $f(x) = \frac{1}{\sqrt{1 - x^2}}$.

To Find:

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

For the function $f(x)$ to be defined as a real number, two conditions must be met:

1. The expression under the square root must be non-negative.

2. The denominator cannot be equal to zero.

From condition 1, the expression inside the square root, $1 - x^2$, must be greater than or equal to zero:

$1 - x^2 \geq 0$

Rearrange the inequality:

$1 \geq x^2$

Or, $x^2 \leq 1$

Taking the square root of both sides gives:

$\sqrt{x^2} \leq \sqrt{1}$

$|x| \leq 1$

The inequality $|x| \leq k$ for $k > 0$ is equivalent to $-k \leq x \leq k$.

In this case, $k=1$. So, the inequality $|x| \leq 1$ is equivalent to:

$-1 \leq x \leq 1$

From condition 2, the denominator $\sqrt{1 - x^2}$ must not be equal to zero.

$\sqrt{1 - x^2} \neq 0$

This implies $1 - x^2 \neq 0$.

We set $1 - x^2 = 0$ to find the values to exclude:

$x^2 = 1$

$x = \pm \sqrt{1}$

$x = \pm 1$

So, we must have $x \neq 1$ and $x \neq -1$.

We need to satisfy both sets of conditions: $-1 \leq x \leq 1$ AND ($x \neq -1$ and $x \neq 1$).

Combining these conditions means that $x$ must be strictly between -1 and 1.

$-1 < x < 1$

In interval notation, the set of all real numbers $x$ such that $-1 < x < 1$ is the open interval $(-1, 1)$.

Thus, the domain of the function $f(x) = \frac{1}{\sqrt{1 - x^2}}$ is $(-1, 1)$.

The final answer is $\boxed{(-1, 1)}$.

Comparing this result with the given options, we find that it matches option (A).

Therefore, the correct option is (A) $(-1, 1)$.

Solution:

Given:

The modulus function $f(x) = |x|$.

To Determine:

The shape of the graph of the modulus function.

The modulus function is defined as $f(x) = |x|$, which means:

$f(x) = x$ if $x \geq 0$

$f(x) = -x$ if $x < 0$

Let's consider the graph by plotting some points:

• If $x=0$, $f(0) = |0| = 0$. Point is $(0, 0)$. (Origin)
• If $x=1$, $f(1) = |1| = 1$. Point is $(1, 1)$.
• If $x=2$, $f(2) = |2| = 2$. Point is $(2, 2)$.
• If $x=-1$, $f(-1) = |-1| = 1$. Point is $(-1, 1)$.
• If $x=-2$, $f(-2) = |-2| = 2$. Point is $(-2, 2)$.

For $x \geq 0$, the graph is $y = x$, which is a straight line passing through the origin with slope 1.

For $x < 0$, the graph is $y = -x$, which is a straight line passing through the origin with slope -1.

When plotted, these two parts form a graph that consists of two rays starting from the origin $(0, 0)$ and extending upwards. The part for $x \geq 0$ goes up and to the right (slope 1), and the part for $x < 0$ goes up and to the left (slope -1). This creates a V-shape with its vertex at the origin.

Let's evaluate the options:

(A) A straight line passing through the origin: This only describes one half of the graph (either $y=x$ or $y=-x$, but not the entire graph of $|x|$). So, this is incorrect.

(B) A V-shaped graph with vertex at the origin, opening upwards: This description perfectly matches the shape formed by the two linear pieces of the modulus function graph.

(C) A parabola opening upwards: A parabola opening upwards has the equation $y = ax^2 + bx + c$ with $a > 0$. For example, $y=x^2$. The graph of $|x|$ is made of straight line segments, not a curve like a parabola. So, this is incorrect.

(D) A horizontal line: A horizontal line has the equation $y = c$ (a constant function). The value of $|x|$ changes with $x$, so it is not a horizontal line (unless restricted to a single point). So, this is incorrect.

Therefore, the graph of the modulus function $f(x) = |x|$ is a V-shaped graph with its vertex at the origin, opening upwards.

The final answer is $\boxed{\text{A V-shaped graph with vertex at the origin, opening upwards}}$.

Therefore, the correct option is (B) A V-shaped graph with vertex at the origin, opening upwards.

Solution:

Given:

The function $f(x)$ is defined as:

$f(x) = \begin{cases} x & , & x \in \mathbb{Q} \\ -x & , & x \notin \mathbb{Q} \end{cases}$

To Find:

The value of $f(\sqrt{2})$.

The definition of the function $f(x)$ depends on whether the input value $x$ is a rational number ($x \in \mathbb{Q}$) or an irrational number ($x \notin \mathbb{Q}$).

We need to evaluate the function at $x = \sqrt{2}$. Therefore, we must determine if $\sqrt{2}$ is a rational number or an irrational number.

A rational number is any number that can be expressed as the quotient or fraction $\frac{p}{q}$ of two integers, a numerator $p$ and a non-zero denominator $q$. An irrational number is a real number that cannot be expressed as a simple fraction.

It is a fundamental result in mathematics that $\sqrt{2}$ cannot be expressed as a fraction of two integers, and thus $\sqrt{2}$ is an irrational number.

This means $\sqrt{2} \notin \mathbb{Q}$.

According to the definition of the function $f(x)$, if $x \notin \mathbb{Q}$ (i.e., $x$ is irrational), then $f(x) = -x$.

Since $\sqrt{2}$ is irrational, we use the second case of the function definition:

$f(x) = -x$ when $x \notin \mathbb{Q}$

Substitute $x = \sqrt{2}$ into this part of the definition:

$f(\sqrt{2}) = -(\sqrt{2})$

$f(\sqrt{2}) = -\sqrt{2}$

Thus, the value of $f(\sqrt{2})$ is $-\sqrt{2}$.

The final answer is $\boxed{-\sqrt{2}}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) $-\sqrt{2}$.

Solution:

Given:

Set $A = \{1, 2\}$.

Set $B = \{a, b, c\}$.

To Determine:

Which of the given options are relations from set $A$ to set $B$. (Multiple correct answers are possible).

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

First, let's find the Cartesian product $A \times B$. This is the set of all ordered pairs $(x, y)$ where the first element $x$ is from set $A$, and the second element $y$ is from set $B$.

$A \times B = \{(x, y) \mid x \in A \text{ and } y \in B\}$

Using the given sets $A = \{1, 2\}$ and $B = \{a, b, c\}$:

$A \times B = \{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)\}$

Now, we check each given option to see if it is a subset of $A \times B$. A set is a subset of another if all its elements are contained within the other set.

(A) $\{(1, a), (2, b)\}$

Let's check each ordered pair in this set:

• $(1, a)$: $1 \in A$ and $a \in B$. So, $(1, a) \in A \times B$.
• $(2, b)$: $2 \in A$ and $b \in B$. So, $(2, b) \in A \times B$.

Since all the ordered pairs in option (A) are elements of $A \times B$, the set in option (A) is a subset of $A \times B$.

Therefore, option (A) is a relation from $A$ to $B$.

(B) $\{(a, 1), (b, 2)\}$

Let's check the ordered pairs:

• $(a, 1)$: $a \in B$ but $a \notin A$. For the pair to be in $A \times B$, the first element must be from $A$. So, $(a, 1) \notin A \times B$.

Since not all elements are in $A \times B$, option (B) is not a relation from $A$ to $B$. (It is a relation from $B$ to $A$).

(C) $\{(1, c), (2, a), (1, b)\}$

Let's check each ordered pair in this set:

• $(1, c)$: $1 \in A$ and $c \in B$. So, $(1, c) \in A \times B$.
• $(2, a)$: $2 \in A$ and $a \in B$. So, $(2, a) \in A \times B$.
• $(1, b)$: $1 \in A$ and $b \in B$. So, $(1, b) \in A \times B$.

Since all the ordered pairs in option (C) are elements of $A \times B$, the set in option (C) is a subset of $A \times B$.

Therefore, option (C) is a relation from $A$ to $B$. (Note that this relation is not a function because the element 1 from set A is related to both c and b in set B).

(D) $\emptyset$

The empty set ($\emptyset$) is considered a subset of every set, including $A \times B$.

Therefore, the empty set is always a relation from $A$ to $B$. This relation is called the empty relation.

Therefore, option (D) is a relation from $A$ to $B$.

Based on the analysis, options (A), (C), and (D) are all subsets of $A \times B$, and thus are relations from $A$ to $B$.

The final answer is $\boxed{\text{(A), (C), (D)}}$.

Therefore, the correct options are (A) $\{(1, a), (2, b)\}$, (C) $\{(1, c), (2, a), (1, b)\}$, and (D) $\emptyset$.

Solution:

Given:

The function $f(x) = x^2 + 2$.

To Find:

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

The range of a function is the set of all possible output values ($f(x)$ or $y$) that the function can produce for inputs in its domain.

The domain of the function $f(x) = x^2 + 2$ is all real numbers, $\mathbb{R}$, since it is a polynomial function.

Let's analyze the expression for $f(x)$. The term $x^2$ is the square of a real number. For any real number $x$, the square $x^2$ is always non-negative (greater than or equal to 0).

$x^2 \geq 0$ for all $x \in \mathbb{R}$

Now, consider the entire expression $f(x) = x^2 + 2$. Since $x^2 \geq 0$, adding 2 to both sides of the inequality gives us the possible values for $f(x)$:

$x^2 + 2 \geq 0 + 2$

$f(x) \geq 2$

This inequality tells us that the value of the function $f(x)$ is always greater than or equal to 2.

The minimum value of $f(x)$ is 2, which occurs when $x^2 = 0$, i.e., when $x=0$.

$f(0) = 0^2 + 2 = 2$.

As $|x|$ increases (either positive or negative), $x^2$ increases, and consequently $f(x) = x^2 + 2$ increases. There is no upper bound for the values of $f(x)$.

Therefore, the set of all possible output values is the set of all real numbers greater than or equal to 2.

In interval notation, the range is $[2, \infty)$.

Comparing this result with the given options, we find that it matches option (B).

The final answer is $\boxed{[2, \infty)}$.

Therefore, the correct option is (B) $[2, \infty)$.

Solution:

Given:

The graph of a relation on $\mathbb{R}$ passes the vertical line test.

To Determine:

What conclusion can be drawn about the relation.

The Vertical Line Test is a graphical method used to determine whether a given graph represents a function. The test states that:

A graph in the Cartesian plane represents a function if and only if no vertical line intersects the graph at more than one point.

If a graph passes the vertical line test, it means that for every value on the x-axis (which represents the input or the first element of the ordered pair), there is at most one corresponding value on the y-axis (which represents the output or the second element of the ordered pair).

Let's relate this to the definition of a function:

A relation is a function if every element in the domain is associated with a unique element in the codomain. In terms of a graph, the domain is represented by the x-values, and the output (image) is represented by the y-values.

Passing the vertical line test means that for each valid x-value, there is only one corresponding y-value. This is exactly the condition required for a relation to be a function.

Let's evaluate the options:

(A) A constant function: A constant function ($f(x) = c$) is a specific type of function. Its graph is a horizontal line. A horizontal line passes the vertical line test. However, passing the vertical line test only guarantees that it is *a* function, not necessarily a constant function (it could be $f(x) = x$, $f(x) = x^2$, etc.). So, this is not the most general conclusion.

(B) An identity function: An identity function ($f(x) = x$) is also a specific type of function. Its graph is a straight line through the origin with slope 1. This graph also passes the vertical line test. However, passing the vertical line test does not mean it *must* be the identity function. So, this is not the most general conclusion.

(C) A relation, but not a function: Since the graph passes the vertical line test, it satisfies the definition of a function. A function is a special type of relation. Therefore, it is a function, which implies it is also a relation. The statement "but not a function" makes this option incorrect.

(D) A function: As explained above, the vertical line test is the graphical criterion for determining if a relation is a function. If the graph of a relation passes the test, the relation is a function.

The most accurate and direct conclusion from a graph passing the vertical line test is that the relation represented by the graph is a function.

The final answer is $\boxed{\text{A function}}$.

Therefore, the correct option is (D) A function.

Solution:

Given:

The function $f = \{(1, 2), (2, 3), (3, 4)\}$.

The domain of the function is $\{1, 2, 3\}$.

The codomain of the function is $\{2, 3, 4, 5\}$.

To Find:

The element in the domain that has the image 3.

A function, when represented as a set of ordered pairs $(x, y)$, relates each element $x$ from the domain to its corresponding image $y$ in the codomain. In the ordered pair $(x, y)$, $x$ is the input (an element from the domain) and $y$ is the output or image of $x$ (an element from the codomain or range).

The given function is $f = \{(1, 2), (2, 3), (3, 4)\}$. Let's examine each ordered pair:

• The ordered pair $(1, 2)$ means that the element 1 from the domain has the image 2. So, $f(1) = 2$.
• The ordered pair $(2, 3)$ means that the element 2 from the domain has the image 3. So, $f(2) = 3$.
• The ordered pair $(3, 4)$ means that the element 3 from the domain has the image 4. So, $f(3) = 4$.

We are looking for the element in the domain that has the image 3. From the ordered pairs, we see that the pair $(2, 3)$ has 3 as the image (the second component).

The corresponding element in the domain (the first component) is 2.

Therefore, the element in the domain $\{1, 2, 3\}$ that has the image 3 is 2.

The final answer is $\boxed{2}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) 2.

Solution:

Given:

Assertion (A): The domain of the function $f(x) = \frac{1}{x}$ is $\mathbb{R} - \{0\}$.

Reason (R): Division by zero is undefined.

Let's evaluate the truthfulness of the Assertion and the Reason, and then check if the Reason correctly explains the Assertion.

Checking Assertion (A):

The function is $f(x) = \frac{1}{x}$. For a real function to be defined, any operation must result in a real number. In this case, the operation is division.

Division is defined for all real numbers except when the divisor (denominator) is zero.

The denominator of $f(x) = \frac{1}{x}$ is $x$. The function is undefined when $x = 0$.

For all other real values of $x$ (i.e., $x \neq 0$), the value of $\frac{1}{x}$ is a well-defined real number.

Therefore, the domain of $f(x) = \frac{1}{x}$ is the set of all real numbers excluding 0, which is $\mathbb{R} - \{0\}$.

Assertion (A) is True.

Checking Reason (R):

Reason (R) states: Division by zero is undefined.

This is a fundamental rule in arithmetic and algebra. In the set of real numbers (and complex numbers), division by zero is an undefined operation.

Reason (R) is True.

Relation between A and R:

Assertion (A) claims the domain of $f(x) = \frac{1}{x}$ is $\mathbb{R} - \{0\}$. Reason (R) states that division by zero is undefined.

The reason why the domain of $f(x) = \frac{1}{x}$ excludes $x=0$ is precisely because division by zero is undefined. The restriction on the domain is a direct consequence of the fact stated in Reason (R).

Therefore, Reason (R) provides the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The final answer is $\boxed{\text{Both A and R are true and R is the correct explanation of A}}$.

Therefore, the correct option is (A) Both A and R are true and R is the correct explanation of A.

Solution:

Given:

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

To Find:

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

The domain of the function $f(x) = \frac{|x|}{x}$ is all real numbers except where the denominator is zero. The denominator is $x$, so the function is defined for all $x \in \mathbb{R}$ such that $x \neq 0$. The domain is $\mathbb{R} - \{0\}$.

Now let's analyze the possible output values for $f(x)$ for $x$ in the domain.

We consider two cases for $x \neq 0$ based on the definition of the absolute value function $|x|$:

Case 1: $x > 0$

If $x > 0$, then $|x| = x$.

So, $f(x) = \frac{|x|}{x} = \frac{x}{x}$.

For any $x > 0$, $\frac{x}{x} = 1$.

Thus, when $x > 0$, the output of the function is always 1.

Case 2: $x < 0$

If $x < 0$, then $|x| = -x$.

So, $f(x) = \frac{|x|}{x} = \frac{-x}{x}$.

For any $x < 0$, $\frac{-x}{x} = -1$.

Thus, when $x < 0$, the output of the function is always -1.

The function is not defined at $x=0$. Therefore, the only possible output values for $f(x)$ are 1 (when $x>0$) and -1 (when $x<0$).

The set of all possible output values is $\{-1, 1\}$.

Note that this function is also known as the Signum function, $\text{sgn}(x)$, excluding the value at $x=0$. The standard Signum function includes $f(0)=0$ in its definition.

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

The final answer is $\boxed{\{-1, 1\}}$.

Comparing this result with the given options, we find that it matches option (A).

Therefore, the correct option is (A) $\{-1, 1\}$.

Solution:

Given:

The function $f(x) = x^3$.

To Find:

The value of the expression $\frac{f(2) - f(1)}{2 - 1}$.

First, we need to evaluate the function $f(x)$ at $x=2$ and $x=1$.

For $f(2)$, substitute $x=2$ into $f(x) = x^3$:

$f(2) = (2)^3$

$f(2) = 2 \times 2 \times 2$

$f(2) = 8$

For $f(1)$, substitute $x=1$ into $f(x) = x^3$:

$f(1) = (1)^3$

$f(1) = 1 \times 1 \times 1$

$f(1) = 1$

Now, substitute the values of $f(2)$ and $f(1)$ into the given expression $\frac{f(2) - f(1)}{2 - 1}$.

$\frac{f(2) - f(1)}{2 - 1} = \frac{8 - 1}{2 - 1}$

Perform the subtraction in the numerator and the denominator:

Numerator: $8 - 1 = 7$

Denominator: $2 - 1 = 1$

Substitute these values back into the fraction:

$\frac{7}{1} = 7$

Thus, the value of $\frac{f(2) - f(1)}{2 - 1}$ is 7.

The final answer is $\boxed{7}$.

Comparing this result with the given options, we find that it matches option (B).

Therefore, the correct option is (B) 7.

Solution:

Given:

Four different functions.

To Determine:

Which of the given functions is a rational function.

A rational function is a function that can be written as the ratio of two polynomial functions, $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials and $Q(x)$ is not the zero polynomial.

Let's examine each option:

(A) $f(x) = x^2 + \sqrt{x}$

This function involves the term $\sqrt{x}$, which is $x^{1/2}$. A polynomial function can only have non-negative integer exponents for the variable. Since the exponent is $\frac{1}{2}$, $\sqrt{x}$ is not a polynomial term (over the domain where $\sqrt{x}$ is defined for real numbers). Therefore, $x^2 + \sqrt{x}$ is not a ratio of polynomials.

This is not a rational function.

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

The numerator is $P(x) = x^2+1$, which is a polynomial.

The denominator is $Q(x) = x-3$, which is a polynomial and is not the zero polynomial.

This function is expressed as the ratio of two polynomials.

Therefore, this is a rational function.

(C) $f(x) = |x| + 5$

This function involves the absolute value function $|x|$. The absolute value function is defined piecewise and is not a polynomial function for all real numbers. For example, its graph is V-shaped, while a polynomial graph is a smooth curve. Therefore, $|x| + 5$ is not a ratio of polynomials.

This is not a rational function.

(D) $f(x) = 2^x$

This function is an exponential function where the variable $x$ is in the exponent. This is not a polynomial function, nor can it generally be expressed as a ratio of polynomials.

This is not a rational function.

Based on the definition, only option (B) fits the criteria of a rational function.

The final answer is $\boxed{f(x) = \frac{x^2+1}{x-3}}$.

Therefore, the correct option is (B) $f(x) = \frac{x^2+1}{x-3}$.

Solution:

Given:

The linear function $f(x) = ax+b$.

The function satisfies $f(1) = 5$ and $f(2) = 8$.

To Find:

The values of the constants $a$ and $b$.

We are given two conditions based on the function's value at specific points. We can substitute these points into the function's equation to form a system of linear equations in terms of $a$ and $b$.

Using the condition $f(1) = 5$:

Substitute $x=1$ into $f(x) = ax+b$:

$f(1) = a(1) + b$

Since $f(1) = 5$, we have:

$a + b = 5$

Using the condition $f(2) = 8$:

Substitute $x=2$ into $f(x) = ax+b$:

$f(2) = a(2) + b$

Since $f(2) = 8$, we have:

$2a + b = 8$

Now we have a system of two linear equations with two variables $a$ and $b$:

1) $a + b = 5$

2) $2a + b = 8$

We can solve this system using the elimination method. Subtract equation (i) from equation (ii):

$(2a + b) - (a + b) = 8 - 5$

$2a + b - a - b = 3$

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

$a + 0 = 3$

$a = 3$

Now that we have the value of $a$, we can substitute $a=3$ into either equation (i) or equation (ii) to find the value of $b$. Using equation (i):

$a + b = 5$

$3 + b = 5$

Solving for $b$:

$b = 5 - 3$

$b = 2$

So, the values of $a$ and $b$ are $a=3$ and $b=2$.

Let's verify these values using equation (ii):

$2a + b = 2(3) + 2 = 6 + 2 = 8$. This matches the given condition $f(2)=8$. The values are correct.

Thus, the values of $a$ and $b$ are 3 and 2, respectively.

The final answer is $\boxed{a=3, b=2}$.

Comparing this result with the given options, we find that it matches option (A).

Therefore, the correct option is (A) $a=3, b=2$.

Solution:

Given:

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

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

Relation $R = \{(1, 2), (1, 3), (2, 4), (3, 5)\}$ from $A$ to $B$.

To Determine:

Which of the given statements about the relation $R$ are correct.

Let's analyze the relation $R$ and evaluate each statement.

$R = \{(1, 2), (1, 3), (2, 4), (3, 5)\}$.

(A) $R$ is a function.

A relation $R$ from set $A$ to set $B$ is a function if and only if every element in set $A$ has one and only one image in set $B$ under the relation $R$. This means that for every element $x \in A$, there must be exactly one ordered pair $(x, y)$ in $R$.

Let's check the elements of set $A = \{1, 2, 3\}$:

• For the element 1: The ordered pairs in $R$ starting with 1 are $(1, 2)$ and $(1, 3)$. The element 1 is related to both 2 and 3 in set $B$. This means 1 has two images (2 and 3).

Since the element 1 from the domain is related to more than one element in the codomain, the relation $R$ is not a function.

Statement (A) is False.

(B) The domain of $R$ is $\{1, 2, 3\}$.

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

The ordered pairs in $R$ are $(1, 2), (1, 3), (2, 4), (3, 5)$.

The first components are 1, 1, 2, and 3.

The set of unique first components is $\{1, 2, 3\}$.

This set is equal to set $A$.

Statement (B) is True.

(C) The range of $R$ is $\{2, 3, 4, 5\}$.

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

The ordered pairs in $R$ are $(1, 2), (1, 3), (2, 4), (3, 5)$.

The second components are 2, 3, 4, and 5.

The set of unique second components is $\{2, 3, 4, 5\}$.

This set is equal to set $B$.

Statement (C) is True.

(D) The element 1 in the domain has unique image.

The image of an element $x$ in the domain under a relation $R$ is the set of all elements $y$ in the codomain such that $(x, y) \in R$. A unique image means this set contains exactly one element.

For the element 1 in the domain, the ordered pairs starting with 1 are $(1, 2)$ and $(1, 3)$.

The set of images of 1 is $\{2, 3\}$.

Since the set $\{2, 3\}$ contains more than one element, the element 1 does not have a unique image.

Statement (D) is False.

Based on the analysis, statements (B) and (C) are correct.

The final answer is $\boxed{\text{(B) and (C)}}$.

Therefore, the correct options are (B) The domain of $R$ is $\{1, 2, 3\}$ and (C) The range of $R$ is $\{2, 3, 4, 5\}$.

Solution:

Given:

Set $A = \{x \in \mathbb{N} : x \leq 3\}$.

Set $B = \{y \in \mathbb{Z} : -1 \leq y \leq 1\}$.

To Find:

The number of elements in the Cartesian product $A \times B$, denoted as $n(A \times B)$.

First, let's determine the elements of set $A$. The set $A$ consists of natural numbers ($x \in \mathbb{N}$) such that $x$ is less than or equal to 3.

Assuming natural numbers start from 1, $\mathbb{N} = \{1, 2, 3, 4, ...\}$.

So, the elements of $A$ that satisfy $x \leq 3$ are 1, 2, and 3.

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

The number of elements in set $A$ is $n(A) = 3$.

Next, let's determine the elements of set $B$. The set $B$ consists of integers ($y \in \mathbb{Z}$) such that $y$ is greater than or equal to -1 and less than or equal to 1.

The integers satisfying $-1 \leq y \leq 1$ are -1, 0, and 1.

$B = \{-1, 0, 1\}$

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

The number of elements in the Cartesian product $A \times B$ is the product of the number of elements in set $A$ and the number of elements in set $B$.

$n(A \times B) = n(A) \times n(B)$

Substitute the calculated values of $n(A)$ and $n(B)$:

$n(A \times B) = 3 \times 3$

$n(A \times B) = 9$

Thus, there are 9 elements in $A \times B$. The elements are the ordered pairs $(x, y)$ where $x \in A$ and $y \in B$.

$A \times B = \{(1, -1), (1, 0), (1, 1), (2, -1), (2, 0), (2, 1), (3, -1), (3, 0), (3, 1)\}$

Comparing this result with the given options, we find that it matches option (B).

The final answer is $\boxed{9}$.

Therefore, the correct option is (B) 9.

Solution:

Given:

The real function $f(x) = \frac{1}{x^2+1}$.

To Find:

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

For the function $f(x)$ to be defined as a real number, the denominator cannot be equal to zero, because division by zero is undefined.

So, we must have:

$x^2 + 1 \neq 0$

Let's try to find the values of $x$ for which the denominator would be zero:

$x^2 + 1 = 0$

Subtract 1 from both sides:

$x^2 = -1$

In the set of real numbers ($\mathbb{R}$), the square of any real number ($x^2$) is always non-negative ($x^2 \geq 0$). Therefore, there is no real number $x$ whose square is equal to -1.

The equation $x^2 = -1$ has no real solutions.

This means that the denominator $x^2 + 1$ is never equal to zero for any real value of $x$. Since the denominator is never zero, the function $f(x) = \frac{1}{x^2+1}$ is defined for all real numbers.

Therefore, the domain of the function $f(x) = \frac{1}{x^2+1}$ is the set of all real numbers.

Domain of $f = \mathbb{R}$ or $(-\infty, \infty)$.

Comparing this result with the given options, we find that it matches option (C).

The final answer is $\boxed{\mathbb{R}}$.

Therefore, the correct option is (C) $\mathbb{R}$.

Solution:

Given:

The function $f(x) = \text{cosec}(x)$.

To Find:

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

The cosecant function is defined as the reciprocal of the sine function:

$f(x) = \text{cosec}(x) = \frac{1}{\sin(x)}$

For $f(x)$ to be a real number, the denominator cannot be equal to zero. Therefore, we must find the values of $x$ for which $\sin(x) = 0$ and exclude them from the domain of real numbers $\mathbb{R}$.

The sine function, $\sin(x)$, is equal to zero at integer multiples of $\pi$. That is, the values of $x$ for which $\sin(x) = 0$ are:

$x = n\pi$, where $n$ is any integer ($n \in \mathbb{Z}$).

These are the values of $x$ where the function $\text{cosec}(x)$ is undefined.

The domain of $f(x) = \text{cosec}(x)$ is the set of all real numbers excluding these values.

Domain of $f = \mathbb{R} - \{x \in \mathbb{R} \mid x = n\pi \text{ for some } n \in \mathbb{Z}\}$

Domain of $f = \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$

Comparing this result with the given options, we find that it matches option (B).

The final answer is $\boxed{\mathbb{R} - \{n\pi : n \in \mathbb{Z}\}}$.

Therefore, the correct option is (B) $\mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$.

Solution:

Given:

The function $f(x) = 2x$.

The function $g(x) = x+5$.

To Find:

The composite function $(f \circ g)(x)$.

The composition of two functions $f$ and $g$, denoted by $(f \circ g)$, is defined as:

$(f \circ g)(x) = f(g(x))$

This means we first apply the function $g$ to $x$, and then apply the function $f$ to the result of $g(x)$.

We are given $g(x) = x+5$.

Now, substitute $g(x)$ into the expression for $f(x)$. The expression for $f(x)$ is $2x$. We replace the variable $x$ in $f(x)$ with the entire expression for $g(x)$.

$f(g(x)) = f(x+5)$

Since $f(x) = 2x$, replacing $x$ with $(x+5)$ gives:

$f(x+5) = 2(x+5)$

Distribute the 2:

$f(x+5) = 2x + 10$

Thus, the composite function $(f \circ g)(x)$ is $2x + 10$.

$(f \circ g)(x) = 2x + 10$

Comparing this result with the given options, we find that it matches option (B).

While option (C) $2(x+5)$ is an intermediate step before simplification, the question asks to "find $(f \circ g)(x)$", and the simplified form is usually expected unless otherwise specified. Option (B) provides the simplified form.

The final answer is $\boxed{2x+10}$.

Therefore, the correct option is (B) $2x+10$.

Solution:

Given:

The greatest integer function $f(x) = \lfloor x \rfloor$.

To Find:

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

The range of a function is the set of all possible output values that the function can produce for inputs in its domain.

The greatest integer function $\lfloor x \rfloor$ gives the largest integer less than or equal to $x$. The domain of this function is all real numbers, $\mathbb{R}$.

Let's consider some examples of the output values of $f(x)$ for different real number inputs $x$:

• If $x = 0.5$, $f(0.5) = \lfloor 0.5 \rfloor = 0$.
• If $x = 1.9$, $f(1.9) = \lfloor 1.9 \rfloor = 1$.
• If $x = 3$, $f(3) = \lfloor 3 \rfloor = 3$.
• If $x = -0.7$, $f(-0.7) = \lfloor -0.7 \rfloor = -1$.
• If $x = -2.4$, $f(-2.4) = \lfloor -2.4 \rfloor = -3$.
• If $x = \pi \approx 3.14$, $f(\pi) = \lfloor \pi \rfloor = 3$.
• If $x = -\sqrt{3} \approx -1.73$, $f(-\sqrt{3}) = \lfloor -\sqrt{3} \rfloor = -2$.

From these examples, we can observe that the output value of the greatest integer function is always an integer.

Furthermore, for any integer $n$, we can choose an input value of $x$ such that $\lfloor x \rfloor = n$. For example, if we want the output to be $n$, we can choose any real number $x$ such that $n \leq x < n+1$. For instance, choosing $x=n$ gives $\lfloor n \rfloor = n$.

This means that any integer can be an output value of the function $f(x) = \lfloor x \rfloor$.

Since the output is always an integer, and any integer can be an output, the set of all possible output values (the range) is the set of all integers.

The set of all integers is denoted by $\mathbb{Z}$.

Range of $f = \mathbb{Z}$

Comparing this result with the given options, we find that it matches option (B).

The final answer is $\boxed{\mathbb{Z}}$.

Therefore, the correct option is (B) $\mathbb{Z}$.

Solution:

Given:

Set $A = \{1, 2, 3\}$, so the number of elements in A is $n(A) = 3$.

Set $B = \{a, b\}$, so the number of elements in B is $n(B) = 2$.

To Find:

The number of functions from set $A$ to set $B$.

A function from set $A$ to set $B$ requires that each element in $A$ is mapped to exactly one element in $B$.

Let $m = n(A)$ be the number of elements in set $A$, and $n = n(B)$ be the number of elements in set $B$.

For each element in set $A$, there are $n(B)$ choices for its image in set $B$.

Since there are $n(A)$ elements in set $A$, and the choice of the image for each element is independent, the total number of ways to map the elements of $A$ to the elements of $B$ is the product of the number of choices for each element in $A$.

Number of functions from $A$ to $B = \underbrace{n(B) \times n(B) \times \cdots \times n(B)}_{n(A) \text{ times}}$

Number of functions from $A$ to $B = (n(B))^{n(A)}$

Number of functions from $A$ to $B = n^m$

Substitute the given values $m = n(A) = 3$ and $n = n(B) = 2$:

Number of functions from $A$ to $B = 2^3$

Calculate the value:

$2^3 = 2 \times 2 \times 2 = 8$

Comparing this result with the given options:

(A) $2^3$

(B) $3^2$

(C) $2 \times 3$

(D) $3 \times 2$

The calculated number of functions, $2^3$, matches option (A).

The final answer is $\boxed{2^3}$.

Therefore, the correct option is (A) $2^3$.

Solution:

Given:

Four different functions with their domains and codomains specified.

To Determine:

Which of the given functions is NOT a real function.

A real function is typically defined as a function whose domain is a subset of the set of real numbers ($\mathbb{R}$). Some definitions also require the codomain to be a subset of $\mathbb{R}$ (making it a real-valued function of a real variable). Let's use the standard definition that the domain must be a subset of $\mathbb{R}$.

Let's examine each option based on its domain:

(A) $f: \mathbb{N} \to \mathbb{R}$ defined by $f(x) = x^2$

The domain of this function is $\mathbb{N}$, the set of natural numbers. The set of natural numbers $\mathbb{N}$ is a subset of the set of real numbers $\mathbb{R}$ ($\mathbb{N} \subset \mathbb{R}$).

The codomain is $\mathbb{R}$, which is a subset of $\mathbb{R}$.

This function satisfies the condition for being a real function.

(B) $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \sqrt{x}$

The domain of this function is $\mathbb{R}$, the set of real numbers. The set of real numbers $\mathbb{R}$ is a subset of $\mathbb{R}$ ($\mathbb{R} \subseteq \mathbb{R}$).

The codomain is $\mathbb{R}$, which is a subset of $\mathbb{R}$.

Based solely on the domain being a subset of $\mathbb{R}$, this fits the definition of a real function. However, it's worth noting that the function rule $\sqrt{x}$ does not map all elements of the stated domain $\mathbb{R}$ to the stated codomain $\mathbb{R}$ (e.g., $\sqrt{-1} = i \notin \mathbb{R}$). But the question asks about the function *as defined* by its stated domain and codomain. The domain itself is a subset of $\mathbb{R}$.

(C) $f: \mathbb{C} \to \mathbb{R}$ defined by $f(z) = |z|$

The domain of this function is $\mathbb{C}$, the set of complex numbers. The set of complex numbers $\mathbb{C}$ is not a subset of the set of real numbers $\mathbb{R}$.

The codomain is $\mathbb{R}$, which is a subset of $\mathbb{R}$.

Since the domain of this function is not a subset of $\mathbb{R}$, it does not satisfy the standard definition of a real function.

(D) $f: \mathbb{R} \to \mathbb{Z}$ defined by $f(x) = \lfloor x \rfloor$

The domain of this function is $\mathbb{R}$, the set of real numbers. $\mathbb{R}$ is a subset of $\mathbb{R}$.

The codomain is $\mathbb{Z}$, the set of integers. The set of integers $\mathbb{Z}$ is a subset of the set of real numbers $\mathbb{R}$ ($\mathbb{Z} \subset \mathbb{R}$).

This function satisfies the condition for being a real function (both domain and codomain are subsets of $\mathbb{R}$).

Based on the standard definition that a real function must have a domain which is a subset of $\mathbb{R}$, option (C) is the only function whose domain is not a subset of $\mathbb{R}$.

The final answer is $\boxed{f: \mathbb{C} \to \mathbb{R} \text{ defined by } f(z) = |z|}$.

Therefore, the correct option is (C) $f: \mathbb{C} \to \mathbb{R}$ defined by $f(z) = |z|$.

Solution:

Given:

Set $L$, the set of all lines in a plane.

Relation $R$ on $L$ defined by $l_1 R l_2$ if $l_1$ is perpendicular to $l_2$. We can write this as $l_1 R l_2 \iff l_1 \perp l_2$.

To Determine:

Whether the relation $R$ is Reflexive, Symmetric, Transitive, or an Equivalence relation.

Let's check the properties of the relation $R$:

1. Reflexivity:

A relation $R$ on a set $L$ is reflexive if $(l, l) \in R$ for every $l \in L$.

In this case, we need to check if every line $l$ is perpendicular to itself ($l \perp l$).

A line is parallel to itself, not perpendicular. The angle a line makes with itself is $0^\circ$, not $90^\circ$.

Thus, $l \not\perp l$ for any line $l$.

Therefore, the relation $R$ is not reflexive.

2. Symmetry:

A relation $R$ on a set $L$ is symmetric if whenever $(l_1, l_2) \in R$, then $(l_2, l_1) \in R$.

In this case, we need to check if whenever $l_1 \perp l_2$, it implies $l_2 \perp l_1$.

If line $l_1$ is perpendicular to line $l_2$, it means the angle between them is $90^\circ$. The statement "$l_2$ is perpendicular to $l_1$" also means the angle between them is $90^\circ$. These two statements are equivalent.

Thus, $l_1 \perp l_2 \implies l_2 \perp l_1$.

Therefore, the relation $R$ is symmetric.

3. Transitivity:

A relation $R$ on a set $L$ is transitive if whenever $(l_1, l_2) \in R$ and $(l_2, l_3) \in R$, then $(l_1, l_3) \in R$.

In this case, we need to check if whenever $l_1 \perp l_2$ and $l_2 \perp l_3$, it implies $l_1 \perp l_3$.

Consider three distinct lines in a plane. If $l_1 \perp l_2$ and $l_2 \perp l_3$, then line $l_1$ must be parallel to line $l_3$. For example, the x-axis ($l_1$) is perpendicular to the y-axis ($l_2$), and the line $x=5$ ($l_3$) is perpendicular to the y-axis ($l_2$). The x-axis ($l_1$) and the line $x=5$ ($l_3$) are parallel, not perpendicular.

Thus, $l_1 \perp l_2$ and $l_2 \perp l_3$ does not imply $l_1 \perp l_3$.

Therefore, the relation $R$ is not transitive.

4. Equivalence Relation:

A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Since $R$ is not reflexive and not transitive, it is not an equivalence relation.

Based on the analysis of the properties, the relation $R$ is symmetric, but not reflexive and not transitive.

The final answer is $\boxed{\text{Symmetric}}$.

Therefore, the correct option is (B) Symmetric.

Solution:

Given:

The function $f(x) = x^2$.

The function $g(x) = 5x$.

To Find:

The value(s) of $x$ for which $f(x) = g(x)$.

We are looking for the values of $x$ where the output of the function $f(x)$ is equal to the output of the function $g(x)$. Set the two function expressions equal to each other:

$f(x) = g(x)$

$x^2 = 5x$

To solve this equation, we move all terms to one side to get a quadratic equation:

$x^2 - 5x = 0$

Factor out the common term $x$ from the expression on the left side:

$x(x - 5) = 0$

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

So, either the first factor is zero or the second factor is zero:

Case 1: $x = 0$

Case 2: $x - 5 = 0$

Solving for $x$ in Case 2:

$x = 5$

Thus, the values of $x$ for which $f(x) = g(x)$ are $x=0$ and $x=5$.

We can check these values:

• If $x=0$: $f(0) = 0^2 = 0$. $g(0) = 5(0) = 0$. $f(0) = g(0)$.
• If $x=5$: $f(5) = 5^2 = 25$. $g(5) = 5(5) = 25$. $f(5) = g(5)$.

The values $x=0$ and $x=5$ satisfy the condition $f(x) = g(x)$.

Comparing this result with the given options, we find that it matches option (C).

The final answer is $\boxed{x=0 \text{ or } x=5}$.

Therefore, the correct option is (C) $x=0$ or $x=5$.

Solution:

Given:

Set $A = \{x, y, z\}$.

To Find:

The number of relations on set $A$.

A relation on set $A$ is defined as any relation from set $A$ to set $A$. In other words, it is any subset of the Cartesian product $A \times A$.

First, find the number of elements in set $A$.

$n(A) = 3$.

Next, find the number of elements in the Cartesian product $A \times A$. If a set has $m$ elements, the number of elements in its Cartesian product with itself is $m \times m = m^2$.

$n(A \times A) = n(A) \times n(A) = 3 \times 3 = 9$.

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

$A \times A = \{(x, x), (x, y), (x, z), (y, x), (y, y), (y, z), (z, x), (z, y), (z, z)\}$

Indeed, there are 9 elements in $A \times A$.

The number of relations on $A$ is equal to the number of possible subsets of $A \times A$. The number of subsets of a set with $k$ elements is $2^k$.

Number of relations on $A = \text{Number of subsets of } A \times A = 2^{n(A \times A)}$

Substitute the calculated value of $n(A \times A)$:

Number of relations = $2^9$

Let's evaluate the options based on this result:

(A) $3^3 = 27$

(B) $2^3 = 8$

(C) $2^{3 \times 3} = 2^9$

(D) $2^6 = 64$

The calculated number of relations, $2^9$, matches option (C) because $3 \times 3 = 9$.

The final answer is $\boxed{2^{3 \times 3}}$.

Therefore, the correct option is (C) $2^{3 \times 3}$.

Solution:

Given:

The constant function $f(x) = c$, where $c$ is a constant.

To Determine:

The shape of the graph of the constant function $f(x) = c$.

The graph of a function $f(x)$ is the set of all points $(x, y)$ in the Cartesian plane such that $y = f(x)$.

For the constant function $f(x) = c$, the equation of the graph is $y = c$.

The equation $y = c$ means that for any value of $x$ in the domain of the function (which is typically $\mathbb{R}$ for a constant function), the corresponding $y$-coordinate is always the same constant value $c$.

Let's consider some points on the graph:

• If $x=0$, $y=c$. Point is $(0, c)$.
• If $x=1$, $y=c$. Point is $(1, c)$.
• If $x=-5$, $y=c$. Point is $(-5, c)$.
• If $x=100$, $y=c$. Point is $(100, c)$.

When plotted on a graph, all these points lie on a straight line where the y-coordinate is always $c$. This line is parallel to the x-axis and passes through the point $(0, c)$ on the y-axis.

A line that is parallel to the x-axis is called a horizontal line.

Let's evaluate the options:

(A) A horizontal line: This matches our conclusion.

(B) A vertical line: A vertical line has an equation of the form $x = k$ (where $k$ is a constant). This is not the form $y=c$.

(C) A parabola: A parabola is a curved graph, typically represented by a quadratic equation like $y = ax^2 + bx + c$ or $x = ay^2 + by + c$. The graph of $y=c$ is a straight line.

(D) A V-shaped curve: A V-shaped curve is characteristic of the modulus function $f(x) = |x|$, not a constant function.

Therefore, the graph of the constant function $f(x) = c$ is a horizontal line.

The final answer is $\boxed{\text{A horizontal line}}$.

Therefore, the correct option is (A) A horizontal line.

Solution:

Given:

The function $f(x) = 2x-1$.

The domain of the function is the set $\{1, 2, 3\}$.

To Find:

The range of the function $f$ for the given domain.

The range of a function for a specific domain is the set of all output values obtained by applying the function rule to each element in the domain.

The domain is $\{1, 2, 3\}$. We need to find $f(x)$ for each value in this set.

For $x=1$:

$f(1) = 2(1) - 1$

$f(1) = 2 - 1$

$f(1) = 1$

For $x=2$:

$f(2) = 2(2) - 1$

$f(2) = 4 - 1$

$f(2) = 3$

For $x=3$:

$f(3) = 2(3) - 1$

$f(3) = 6 - 1$

$f(3) = 5$

The output values obtained are 1, 3, and 5 when the inputs are 1, 2, and 3 respectively.

The range of the function for the domain $\{1, 2, 3\}$ is the set of these output values.

Range of $f = \{1, 3, 5\}$

Comparing this result with the given options, we find that it matches option (A).

The final answer is $\boxed{\{1, 3, 5\}}$.

Therefore, the correct option is (A) $\{1, 3, 5\}$.

Solution:

To Determine:

Which of the given options is a representation of a relation.

A relation is a set of ordered pairs that describes a relationship between elements of two sets (or elements within the same set).

Let's examine each option:

(A) Set-builder form:

A relation can be described using set-builder notation, which specifies the rule or conditions that define the ordered pairs in the relation.

Example: $R = \{(x, y) : x \in A, y \in B, y = x+1\}$. This form describes the set of ordered pairs $(x, y)$ that satisfy the given condition.

Therefore, set-builder form is a representation of a relation.

(B) Roster form:

A relation can be listed explicitly by enumerating all the ordered pairs that belong to the relation. This is known as the roster form.

Example: If $A = \{1, 2\}$ and $B = \{2, 3\}$, the relation $R = \{(1, 2), (2, 3)\}$ lists the specific ordered pairs.

Therefore, roster form is a representation of a relation.

(C) Arrow diagram:

An arrow diagram is a visual representation of a relation. It involves drawing ovals or regions to represent the sets, and drawing arrows from an element in the first set to an element in the second set if they are related by the given relation.

Example: For the relation $R = \{(1, 2), (2, 3)\}$ from $A$ to $B$, we draw an arrow from 1 in $A$ to 2 in $B$, and an arrow from 2 in $A$ to 3 in $B$.

Therefore, an arrow diagram is a representation of a relation.

Since set-builder form, roster form, and arrow diagrams are all valid ways to represent a relation, the option that includes all of them is correct.

The final answer is $\boxed{\text{All of the above}}$.

Therefore, the correct option is (D) All of the above.

Solution:

Given:

The function $f(x) = \sin(x)$.

To Find:

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

The range of a function is the set of all possible output values that the function can produce for inputs in its domain.

The domain of the sine function $f(x) = \sin(x)$ is all real numbers, $\mathbb{R}$.

The sine function is a periodic function that oscillates between a minimum value and a maximum value.

The maximum value that $\sin(x)$ can attain is 1.

The minimum value that $\sin(x)$ can attain is -1.

For any real number $x$, the value of $\sin(x)$ is always between -1 and 1, inclusive.

So, for all $x \in \mathbb{R}$, we have:

$-1 \leq \sin(x) \leq 1$

The set of all possible output values of the function $f(x) = \sin(x)$ is the set of all real numbers $y$ such that $-1 \leq y \leq 1$.

In interval notation, this set is represented by the closed interval $[-1, 1]$.

Range of $f = [-1, 1]$

Comparing this result with the given options:

(A) $(-1, 1)$ - This is an open interval, excluding -1 and 1.

(B) $[-1, 1]$ - This is a closed interval, including -1 and 1.

(C) $\mathbb{R}$ - This is the set of all real numbers, which is incorrect as the sine function's output is bounded.

(D) $[0, \infty)$ - This is the set of non-negative real numbers, which is incorrect as the sine function can output negative values.

The final answer is $\boxed{[-1, 1]}$.

Therefore, the correct option is (B) $[-1, 1]$.

Solution:

Given:

The function $f(x) = 5$ for all $x \in \mathbb{R}$.

To Determine:

Which type of function is $f(x) = 5$.

Let's analyze the properties and classifications of functions to match the given function $f(x) = 5$.

A Constant function is a function whose output value is the same for every input value. The general form is $f(x) = c$, where $c$ is a constant.

The given function $f(x) = 5$ fits this definition exactly, with the constant $c = 5$. The output is always 5, regardless of the input $x$.

A Polynomial function is a function that can be expressed in the form $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $a_0, a_1, \dots, a_n$ are constants and $n$ is a non-negative integer (the degree of the polynomial).

The given function $f(x) = 5$ can be written as $f(x) = 5 \cdot x^0$. This fits the general form of a polynomial where $n=0$ and $a_0 = 5$. A constant function (that is not the zero function $f(x)=0$) is a polynomial of degree zero.

An Identity function is a function that always returns the input value as the output value. The general form is $f(x) = x$.

The given function $f(x) = 5$ is not an identity function because $f(x)$ is always 5, whereas the identity function's output changes with the input $x$ (e.g., $f(1)=1$, $f(5)=5$).

Based on the analysis, the function $f(x) = 5$ is both a constant function and a polynomial function (specifically, a polynomial of degree 0).

Let's evaluate the given options:

(A) Identity function: Incorrect, as $f(x) \neq x$.

(B) Constant function: Correct, as $f(x)$ has a constant value 5 for all $x$.

(C) Polynomial function: Correct, as $f(x) = 5$ is a polynomial of degree 0.

(D) Both (B) and (C): This option combines the two correct classifications identified above.

Since the function $f(x)=5$ is indeed a constant function and a polynomial function, option (D) is the most comprehensive and correct answer among the choices.

The final answer is $\boxed{\text{Both (B) and (C)}}$.

Therefore, the correct option is (D) Both (B) and (C).

Solution:

Given:

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

The relation $R$ on $A$ is defined as $R = \{(x, y) : x, y \in A, \text{ and } x \text{ divides } y\}$.

To Determine:

Which of the given ordered pairs is NOT in the relation $R$.

According to the definition of the relation $R$, an ordered pair $(x, y)$ is in $R$ if and only if both $x$ and $y$ are elements of the set $A = \{1, 2, 3, 4\}$, and $x$ is a divisor of $y$ (or $y$ is a multiple of $x$).

Let's list the elements of the relation $R$ by considering all possible pairs $(x, y)$ where $x, y \in A$ and checking if $x$ divides $y$:

• For $x=1$: 1 divides 1, 2, 3, 4. Pairs are $(1, 1), (1, 2), (1, 3), (1, 4)$.
• For $x=2$: 2 divides 2, 4. Pairs are $(2, 2), (2, 4)$.
• For $x=3$: 3 divides 3. Pair is $(3, 3)$. (3 does not divide 1, 2, or 4 in A)
• For $x=4$: 4 divides 4. Pair is $(4, 4)$. (4 does not divide 1, 2, or 3 in A)

The relation $R$ in roster form is:

$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)\}$

Now, we check each of the given options to see if it is present in the set $R$ or not.

(A) $(1, 4)$: Is $(1, 4) \in R$? Yes, $1 \in A$, $4 \in A$, and 1 divides 4. So, $(1, 4)$ is in $R$.

(B) $(2, 3)$: Is $(2, 3) \in R$? $2 \in A$ and $3 \in A$. Does 2 divide 3? No, 3 is not an integer multiple of 2.

So, $(2, 3)$ is NOT in $R$.

(C) $(3, 3)$: Is $(3, 3) \in R$? Yes, $3 \in A$, $3 \in A$, and 3 divides 3. So, $(3, 3)$ is in $R$.

(D) $(4, 4)$: Is $(4, 4) \in R$? Yes, $4 \in A$, $4 \in A$, and 4 divides 4. So, $(4, 4)$ is in $R$.

The only pair among the options that is not an element of the relation $R$ is $(2, 3)$.

The final answer is $\boxed{(2, 3)}$.

Therefore, the correct option is (B) $(2, 3)$.

Solution:

Given:

The piecewise function $f(x) = \begin{cases} x+2 & , & x < 0 \\ x^2 & , & x \geq 0 \end{cases}$.

To Find:

The values of $f(-3)$ and $f(3)$.

To evaluate the function at a specific value of $x$, we first need to determine which part of the piecewise definition applies based on the value of $x$.

Find $f(-3)$:**