Chapter 1 Relations And Functions

Given:

Set $A = \{0, 1, 2, 3\}$

Relation $R = \{(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)\}$ on set A.

To Determine:

Whether R is reflexive, symmetric, and transitive.

Solution:

We will check each property for the relation R on the set A.

(i) Reflexivity:

A relation R on a set A is said to be reflexive if $(a, a) \in R$ for every $a \in A$.

In this case, the set is $A = \{0, 1, 2, 3\}$. We need to check if $(0, 0), (1, 1), (2, 2), (3, 3)$ are all in R.

From the given relation R, we observe that:

$(0, 0) \in R$

$(1, 1) \in R$

$(2, 2) \in R$

$(3, 3) \in R$

Since $(a, a) \in R$ for all $a \in A$, the relation R is reflexive.

(ii) Symmetry:

A relation R on a set A is said to be symmetric if for all $a, b \in A$, whenever $(a, b) \in R$, then $(b, a)$ must also be in R.

We check the pairs in R:

For $(0, 0) \in R$, we need $(0, 0) \in R$. This is true.

For $(0, 1) \in R$, we need $(1, 0) \in R$. From the given R, $(1, 0) \in R$. This is true.

For $(0, 3) \in R$, we need $(3, 0) \in R$. From the given R, $(3, 0) \in R$. This is true.

For $(1, 0) \in R$, we need $(0, 1) \in R$. From the given R, $(0, 1) \in R$. This is true. (Already checked)

For $(1, 1) \in R$, we need $(1, 1) \in R$. This is true.

For $(2, 2) \in R$, we need $(2, 2) \in R$. This is true.

For $(3, 0) \in R$, we need $(0, 3) \in R$. From the given R, $(0, 3) \in R$. This is true. (Already checked)

For $(3, 3) \in R$, we need $(3, 3) \in R$. This is true.

Since for every $(a, b) \in R$, $(b, a)$ is also in R, the relation R is symmetric.

(iii) Transitivity:

A relation R on a set A is said to be transitive if for all $a, b, c \in A$, whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c)$ must also be in R.

We check pairs $(a, b) \in R$ and $(b, c) \in R$ to see if $(a, c) \in R$.

Let's list relevant combinations:

$(0, 1) \in R$ and $(1, 0) \in R$. We need $(0, 0) \in R$. Yes, $(0, 0) \in R$.

$(0, 1) \in R$ and $(1, 1) \in R$. We need $(0, 1) \in R$. Yes, $(0, 1) \in R$.

$(0, 3) \in R$ and $(3, 0) \in R$. We need $(0, 0) \in R$. Yes, $(0, 0) \in R$.

$(0, 3) \in R$ and $(3, 3) \in R$. We need $(0, 3) \in R$. Yes, $(0, 3) \in R.

$(1, 0) \in R$ and $(0, 0) \in R$. We need $(1, 0) \in R$. Yes, $(1, 0) \in R.

$(1, 0) \in R$ and $(0, 1) \in R$. We need $(1, 1) \in R$. Yes, $(1, 1) \in R$.

$(1, 0) \in R$ and $(0, 3) \in R$. We need $(1, 3) \in R$. From the given R, $(1, 3) \notin R$.

Since we found a pair $(1, 0) \in R$ and $(0, 3) \in R$ but $(1, 3) \notin R$, the relation R is not transitive.

Conclusion:

Based on the checks:

R is reflexive.

R is symmetric.

R is not transitive.

Given:

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

Relation $R = \{(1, 1), (2, 2), (3, 3), (1, 3)\}$ on set A.

To Find:

The ordered pairs to be added to R to make it the smallest equivalence relation.

Solution:

For R to be an equivalence relation, it must be reflexive, symmetric, and transitive.

(i) Reflexivity:

For R to be reflexive on set $A=\{1, 2, 3\}$, it must contain $(1, 1), (2, 2),$ and $(3, 3)$.

The given relation $R$ already contains $(1, 1), (2, 2),$ and $(3, 3)$.

So, R is already reflexive. No pairs need to be added for reflexivity.

(ii) Symmetry:

For R to be symmetric, if $(a, b) \in R$, then $(b, a)$ must also be in R.

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

We check the non-diagonal pairs:

We have $(1, 3) \in R$. For symmetry, we must have $(3, 1) \in R$.

The pair $(3, 1)$ is not in the original R.

To make R symmetric, we must add $(3, 1)$.

Current state of R for checking transitivity: $R' = \{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}$.

(iii) Transitivity:

For R to be transitive, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c)$ must also be in R.

We check the current relation $R' = \{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}$.

Consider pairs where the second element of one matches the first element of another:

$(1, 3) \in R'$ and $(3, 1) \in R' \implies$ we need $(1, 1) \in R'$. Yes, $(1, 1) \in R'$.

$(3, 1) \in R'$ and $(1, 3) \in R' \implies$ we need $(3, 3) \in R'$. Yes, $(3, 3) \in R'$.

$(1, 1) \in R'$ and $(1, 3) \in R' \implies$ we need $(1, 3) \in R'$. Yes, $(1, 3) \in R'$.

$(3, 3) \in R'$ and $(3, 1) \in R' \implies$ we need $(3, 1) \in R'$. Yes, $(3, 1) \in R'$.

All required transitive pairs are present in $R'$. No further pairs need to be added for transitivity after adding $(3, 1)$ for symmetry.

Conclusion:

To make R reflexive, we need no additions.

To make R symmetric, we need to add $(3, 1)$.

Adding $(3, 1)$ makes the relation transitive as well.

The smallest set of ordered pairs that must be added to R to make it an equivalence relation is $\{(3, 1)\}$.

The smallest equivalence relation containing the given R is $R'' = \{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}$.

Given:

Set $Z$ of integers.

Equivalence relation $R = \{(a, b) : 2 \text{ divides } a - b\}$ on $Z$.

To Find:

The equivalence class of 0, denoted by [0].

Solution:

The equivalence class of an element $a \in Z$ with respect to an equivalence relation R is defined as the set of all elements $x \in Z$ such that $(x, a) \in R$. It is denoted by $[a]$.

In this problem, we need to find the equivalence class of 0, which is $[0]$.

By definition,

$[0] = \{x \in Z : (x, 0) \in R\}$.

The given relation R is defined by $(a, b) \in R$ if and only if 2 divides $a - b$.

Applying this definition to the pair $(x, 0)$, we have $(x, 0) \in R$ if and only if 2 divides $x - 0$.

This means 2 divides $x$.

An integer $x$ is divisible by 2 if and only if $x$ is an even integer.

Therefore, the equivalence class [0] consists of all integers that are divisible by 2.

So, $[0]$ is the set of all even integers.

$[0] = \{x \in Z : x \text{ is an even integer}\}$.

We can also list some elements of this set:

$[0] = \{..., -6, -4, -2, 0, 2, 4, 6, ...\}$

Given:

A function $f : R \to R$ defined by $f(x) = 4x - 1$ for all $x \in R$.

To Show:

The function f is one-one (or injective).

Solution:

A function $f: A \to B$ is defined as one-one if for any two elements $x_1, x_2$ in the domain A, if $f(x_1) = f(x_2)$, then it must follow that $x_1 = x_2$.

Let $x_1, x_2 \in R$ be two elements in the domain such that $f(x_1) = f(x_2)$.

Using the definition of the function $f(x) = 4x - 1$, we can write:

To isolate the terms involving $x_1$ and $x_2$, we add 1 to both sides of the equation:

Simplifying both sides, we get:

Now, divide both sides of the equation by 4 (which is a non-zero constant):

This simplifies to:

We assumed $f(x_1) = f(x_2)$ and through algebraic steps, we derived that $x_1 = x_2$. This matches the definition of a one-one function.

Conclusion:

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in R$, the function $f(x) = 4x - 1$ is one-one.

Given:

Function $f = \{(5, 2), (6, 3)\}$

Function $g = \{(2, 5), (3, 6)\}$

To Find:

The composite function $f \circ g$.

Solution:

The composition of functions $f$ and $g$, denoted by $f \circ g$, is defined as $(f \circ g)(x) = f(g(x))$.

For an ordered pair $(x, z)$ to be in $f \circ g$, there must exist some element $y$ such that $(x, y) \in g$ and $(y, z) \in f$.

We look at the ordered pairs in $g$ and trace the mapping through $f$.

Consider the first ordered pair in $g$, which is $(2, 5)$.

Here, $x = 2$ and $g(2) = 5$.

Now we need to find the value of $f$ at $g(2)$, which is $f(5)$.

Looking at the ordered pairs in $f$, we see that $(5, 2) \in f$, which means $f(5) = 2$.

So, $(f \circ g)(2) = f(g(2)) = f(5) = 2$.

This gives us the ordered pair $(2, 2)$ for $f \circ g$.

Consider the second ordered pair in $g$, which is $(3, 6)$.

Here, $x = 3$ and $g(3) = 6$.

Now we need to find the value of $f$ at $g(3)$, which is $f(6)$.

Looking at the ordered pairs in $f$, we see that $(6, 3) \in f$, which means $f(6) = 3$.

So, $(f \circ g)(3) = f(g(3)) = f(6) = 3$.

This gives us the ordered pair $(3, 3)$ for $f \circ g$.

We have processed all ordered pairs in $g$.

The set of ordered pairs for $f \circ g$ is the collection of the pairs we found.

$f \circ g = \{(2, 2), (3, 3)\}$.

To find the inverse function $f^{-1}$ of the given function $f(x) = 4x - 3$, we follow a standard procedure.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 4x - 3$ for all $x \in \mathbb{R}$.

To Find:

The inverse function $f^{-1}(x)$.

Solution:

Let $y = f(x)$. Then we have:

$y = 4x - 3$

To find the inverse function, we need to express $x$ in terms of $y$. We rearrange the equation:

Add 3 to both sides:

$y + 3 = 4x$

Divide both sides by 4:

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

Now, to get the inverse function $f^{-1}(x)$, we swap $x$ and $y$. So, wherever we see $y$, we replace it with $x$, and the expression for $x$ in terms of $y$ becomes the expression for $f^{-1}(x)$ in terms of $x$.

Thus, replacing $y$ with $x$ on the right side, we get the inverse function:

$f^{-1}(x) = \frac{x + 3}{4}$

Final Answer:

The inverse function of $f(x) = 4x - 3$ is $\boxed{f^{-1}(x) = \frac{x + 3}{4}}$.

A binary operation $*$ defined on a set $S$ is said to be commutative if for every pair of elements $a, b \in S$, the property $a * b = b * a$ holds true.

Given:

The set is the set of integers, $\mathbb{Z}$.

The binary operation $*$ is defined as $m * n = m - n + mn$ for all $m, n \in \mathbb{Z}$.

To Check:

Whether the operation $*$ is commutative on $\mathbb{Z}$.

Solution:

For the operation $*$ to be commutative, we must have $m * n = n * m$ for all $m, n \in \mathbb{Z}$.

Let's compute $m * n$ and $n * m$ based on the given definition.

According to the definition:

$m * n = m - n + mn$

Now, let's compute $n * m$. To do this, we swap the roles of $m$ and $n$ in the definition:

$n * m = n - m + nm$

For the operation to be commutative, we need $m * n = n * m$, which means:

$m - n + mn = n - m + nm$

Since multiplication of integers is commutative, $mn = nm$. So, the equation simplifies to:

$m - n + mn = n - m + mn$

Subtracting $mn$ from both sides, we get:

$m - n = n - m$

Rearranging the terms to one side:

$m - n - (n - m) = 0$

$m - n - n + m = 0$

$2m - 2n = 0$

$2(m - n) = 0$

This equation $2(m - n) = 0$ is true if and only if $m - n = 0$, which means $m = n$.

The condition $m * n = n * m$ is therefore only true when $m = n$. It is not true for all $m, n \in \mathbb{Z}$ where $m \neq n$.

Let's take a specific example to demonstrate this. Let $m = 1$ and $n = 2$. Both $1, 2 \in \mathbb{Z}$.

Compute $1 * 2$:

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

Compute $2 * 1$:

$2 * 1 = 2 - 1 + (2)(1) = 1 + 2 = 3$

Since $1 * 2 = 1$ and $2 * 1 = 3$, we see that $1 * 2 \neq 2 * 1$.

Because we found at least one pair of integers $(1, 2)$ for which $m * n \neq n * m$, the operation $*$ is not commutative on $\mathbb{Z}$.

Conclusion:

The binary operation $*$ defined by $m * n = m - n + mn$ on the set of integers $\mathbb{Z}$ is not commutative.

The range of a function represented as a set of ordered pairs $(a, b)$ is the set of all the second elements (the $b$ values) from the ordered pairs.

Given:

Function $f$ is given as a set of ordered pairs: $f = \{(5, 2), (6, 3)\}$.

Function $g$ is given as a set of ordered pairs: $g = \{(2, 5), (3, 6)\}$.

To Find:

The range of function $f$ and the range of function $g$.

Solution:

For function $f = \{(5, 2), (6, 3)\}$, the ordered pairs are $(5, 2)$ and $(6, 3)$.

The second elements in these pairs are 2 and 3.

Therefore, the range of $f$ is the set containing these second elements.

Range of $f = \{2, 3\}$

For function $g = \{(2, 5), (3, 6)\}$, the ordered pairs are $(2, 5)$ and $(3, 6)$.

The second elements in these pairs are 5 and 6.

Therefore, the range of $g$ is the set containing these second elements.

Range of $g = \{5, 6\}$

Final Answer:

The range of $f$ is $\{2, 3\}$.

The range of $g$ is $\{5, 6\}$.

A relation from set A to set B is called a function if every element of set A has one and only one image in set B. In terms of ordered pairs, a relation is a function if no two distinct ordered pairs have the same first element.

Given:

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

Relation $f = \{(1, 3), (2, 3), (3, 2)\}$, which is a subset of $A \times A$.

Relation $g = \{(1, 2), (1, 3), (3, 1)\}$, which is a subset of $A \times A$.

To Determine:

Which of the relations, $f$ or $g$, is a function, and provide the reason.

Solution:

Let's examine relation $f = \{(1, 3), (2, 3), (3, 2)\}$.

The domain of $f$ is the set of all first elements in the ordered pairs, which is $\{1, 2, 3\}$. This set is equal to set A.

Now let's check if each element in the domain is associated with exactly one element in the codomain (which is also A).

• The element 1 from the domain is associated with 3 (from the pair (1, 3)).
• The element 2 from the domain is associated with 3 (from the pair (2, 3)).
• The element 3 from the domain is associated with 2 (from the pair (3, 2)).

Every element in set A ($\{1, 2, 3\}$) appears exactly once as the first element of an ordered pair in $f$. Therefore, no two distinct ordered pairs in $f$ have the same first element.

Thus, relation $f$ satisfies the condition for being a function.

Now let's examine relation $g = \{(1, 2), (1, 3), (3, 1)\}$.

The domain of $g$ is the set of all first elements in the ordered pairs, which is $\{1, 3\}$. This set is a subset of A, but not equal to A. (Note: While a relation might not cover the entire domain, the primary check for a function is whether any element from the domain is mapped to multiple elements in the codomain).

Let's check if any first element is associated with more than one second element.

• The element 1 from the domain is associated with 2 (from the pair (1, 2)).
• The element 1 from the domain is also associated with 3 (from the pair (1, 3)).

Here, the element 1 from the set A is associated with two different elements (2 and 3) in the codomain (set A). This violates the definition of a function, which requires that each element in the domain must have one and only one image.

The ordered pairs $(1, 2)$ and $(1, 3)$ show that the first element 1 is mapped to two different second elements, 2 and 3. Therefore, relation $g$ is not a function.

Conclusion:

Relation f is a function because every element in the set A ($\{1, 2, 3\}$) appears exactly once as the first element in the ordered pairs of $f$.

Relation g is not a function because the element 1 from set A appears as the first element in two different ordered pairs, $(1, 2)$ and $(1, 3)$, meaning it is associated with two distinct elements (2 and 3) in the codomain.

Given: Set $A = \{a, b, c, d\}$ and function $f = \{(a, b), (b, d), (c, a), (d, c)\}$.

To Show: $f$ is one-one and onto from A to A. To Find: $f^{-1}$.

Solution:

For function $f$, the images of the elements in A are:

• $f(a) = b$
• $f(b) = d$
• $f(c) = a$
• $f(d) = c$

One-one: The images $\{b, d, a, c\}$ are all distinct elements in the codomain A. Since distinct elements in the domain A have distinct images in the codomain A, $f$ is one-one.

Onto: The range of $f$ is the set of all second elements in the ordered pairs, which is $\{b, d, a, c\}$. This set is equal to the codomain A ($\{a, b, c, d\}$). Since the range equals the codomain, $f$ is onto.

Since $f$ is both one-one and onto, it is a bijection, and its inverse $f^{-1}$ exists.

Finding f^–1: To find the inverse function, we swap the elements in each ordered pair of $f$.

$f = \{(a, b), (b, d), (c, a), (d, c)\}$

Swapping the pairs, we get:

$f^{-1} = \{(b, a), (d, b), (a, c), (c, d)\}$

Or, written with domain elements sorted: $f^{-1} = \{(a, c), (b, a), (c, d), (d, b)\}$.

Final Answer:

$f$ is one-one as $f(x_1) = f(x_2) \implies x_1 = x_2$.

$f$ is onto as Range($f$) = Codomain($A$).

$f^{-1} = \{(a, c), (b, a), (c, d), (d, b)\}$.

A binary operation $*$ defined on a set $S$ is commutative if $a * b = b * a$ for all $a, b \in S$.

A binary operation $*$ defined on a set $S$ is associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in S$.

Given:

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

The binary operation $*$ is defined as $m * n = \text{g.c.d.}(m, n)$ for all $m, n \in \mathbb{N}$. (g.c.d. stands for greatest common divisor).

To Check:

Whether the operation $*$ is commutative and associative on $\mathbb{N}$.

Solution:

Checking for Commutativity:

We need to check if $m * n = n * m$ for all $m, n \in \mathbb{N}$.

By definition, $m * n = \text{g.c.d.}(m, n)$.

And $n * m = \text{g.c.d.}(n, m)$.

The greatest common divisor of two natural numbers is independent of the order of the numbers. That is, $\text{g.c.d.}(m, n) = \text{g.c.d.}(n, m)$ for all $m, n \in \mathbb{N}$.

For example, $6 * 9 = \text{g.c.d.}(6, 9) = 3$, and $9 * 6 = \text{g.c.d.}(9, 6) = 3$. So, $6 * 9 = 9 * 6$.

Since $\text{g.c.d.}(m, n) = \text{g.c.d.}(n, m)$ for all $m, n \in \mathbb{N}$, the operation $*$ is commutative.

Checking for Associativity:

We need to check if $(m * n) * p = m * (n * p)$ for all $m, n, p \in \mathbb{N}$.

Let's evaluate the left side:

$(m * n) * p = \text{g.c.d.}(m, n) * p = \text{g.c.d.}(\text{g.c.d.}(m, n), p)$

Let's evaluate the right side:

$m * (n * p) = m * \text{g.c.d.}(n, p) = \text{g.c.d.}(m, \text{g.c.d.}(n, p))$

We need to check if $\text{g.c.d.}(\text{g.c.d.}(m, n), p) = \text{g.c.d.}(m, \text{g.c.d.}(n, p))$.

The greatest common divisor of three natural numbers $m, n, p$, denoted as $\text{g.c.d.}(m, n, p)$, can be computed by finding the g.c.d. of any two numbers and then finding the g.c.d. of the result with the third number. This property holds for g.c.d. (and also for l.c.m.).

That is, $\text{g.c.d.}(\text{g.c.d.}(m, n), p) = \text{g.c.d.}(m, n, p)$ and $\text{g.c.d.}(m, \text{g.c.d.}(n, p)) = \text{g.c.d.}(m, n, p)$.

Since both sides are equal to $\text{g.c.d.}(m, n, p)$, the equation holds true for all $m, n, p \in \mathbb{N}$.

For example, let $m = 12, n = 18, p = 30$.

Left side: $(12 * 18) * 30 = \text{g.c.d.}(12, 18) * 30 = 6 * 30 = \text{g.c.d.}(6, 30) = 6$.

Right side: $12 * (18 * 30) = 12 * \text{g.c.d.}(18, 30) = 12 * 6 = \text{g.c.d.}(12, 6) = 6$.

Since $(12 * 18) * 30 = 12 * (18 * 30)$, the property holds for this example.

Since $\text{g.c.d.}(\text{g.c.d.}(m, n), p) = \text{g.c.d.}(m, \text{g.c.d.}(n, p))$ for all $m, n, p \in \mathbb{N}$, the operation $*$ is associative.

Conclusion:

The binary operation $*$ defined by $m * n = \text{g.c.d.}(m, n)$ on the set of natural numbers $\mathbb{N}$ is both commutative and associative.

Given:

Set $N = \{1, 2, 3, ...\}$ (the set of natural numbers).

Relation $R$ defined on $N$ such that for $n, m \in N$, $nRm$ if and only if the remainder of $n$ on division by 5 is in $\{0, 1, 2, 3, 4\}$ and the remainder of $m$ on division by 5 is in $\{0, 1, 2, 3, 4\}$.

To Show:

R is an equivalence relation.

Also, find the pairwise disjoint subsets determined by R.

Solution:

We are given the set of natural numbers $N = \{1, 2, 3, ...\}$.

The relation R is defined as $nRm$ if and only if $(n \pmod 5 \in \{0, 1, 2, 3, 4\})$ and $(m \pmod 5 \in \{0, 1, 2, 3, 4\})$.

For any natural number $n \in N$, division by 5 yields a remainder $r$ such that $0 \le r < 5$. Thus, the remainder $n \pmod 5$ is always in the set $\{0, 1, 2, 3, 4\}$.

Therefore, the condition $(n \pmod 5 \in \{0, 1, 2, 3, 4\})$ is true for every $n \in N$.

Similarly, the condition $(m \pmod 5 \in \{0, 1, 2, 3, 4\})$ is true for every $m \in N$.

Since both conditions are always true for any $n, m \in N$, the relation $nRm$ holds for all pairs $(n, m) \in N \times N$.

Hence, $R = N \times N$. This is the universal relation on $N$.

To show that R is an equivalence relation, we must verify reflexivity, symmetry, and transitivity.

Reflexivity:

For any $n \in N$, we need to check if $nRn$.

Since $R = N \times N$, the pair $(n, n)$ is always in $R$ for any $n \in N$.

Thus, $nRn$ for all $n \in N$. R is reflexive.

Symmetry:

For any $n, m \in N$, we need to check if $nRm \implies mRn$.

If $nRm$, then $(n, m) \in R$. Since $R = N \times N$, this is true for any $n, m \in N$.

If $(n, m) \in R$, then $(m, n)$ is also in $R$ because $R$ contains all possible pairs from $N \times N$.

Thus, if $nRm$, then $mRn$. R is symmetric.

Transitivity:

For any $n, m, p \in N$, we need to check if $nRm$ and $mRp \implies nRp$.

If $nRm$ and $mRp$, this means $(n, m) \in R$ and $(m, p) \in R$.

Since $R = N \times N$, both $(n, m) \in R$ and $(m, p) \in R$ are true for any $n, m, p \in N$.

Also, since $R = N \times N$, the pair $(n, p)$ is always in $R$ for any $n, p \in N$.

Thus, if $nRm$ and $mRp$, then $nRp$. R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation.

Pairwise Disjoint Subsets determined by R (Equivalence Classes):

The equivalence classes determined by R partition the set N into disjoint subsets.

The equivalence class of an element $x \in N$, denoted by $[x]$, is the set of all elements in N that are related to $x$ by R. That is, $[x] = \{y \in N \mid yRx\}$.

Since $yRx$ is true for all $y \in N$ (because $R = N \times N$), the equivalence class of any element $x \in N$ is the set N itself.

For example, $[1] = \{y \in N \mid yR1\}$. Since $yR1$ is true for all $y \in N$, $[1] = N$.

Similarly, for any $x \in N$, $[x] = N$.

The set of all distinct equivalence classes is $\{N\}$.

The pairwise disjoint subsets determined by R is the single set N. This forms a partition of N into one subset.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{x}{x^2 + 1}$ for all $x \in \mathbb{R}$.

To Show:

The function $f$ is neither one-one (injective) nor onto (surjective).

Proof:

Not One-one:

A function $f$ is one-one if for any $x_1, x_2$ in the domain, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

To show that $f$ is not one-one, we need to find two distinct values $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$.

Let $f(x_1) = f(x_2)$.

$\frac{x_1}{x_1^2 + 1} = \frac{x_2}{x_2^2 + 1}$

$x_1(x_2^2 + 1) = x_2(x_1^2 + 1)$

$x_1 x_2^2 + x_1 = x_2 x_1^2 + x_2$

$x_1 x_2^2 - x_2 x_1^2 + x_1 - x_2 = 0$

$x_1 x_2(x_2 - x_1) - 1(x_2 - x_1) = 0$

$(x_1 x_2 - 1)(x_2 - x_1) = 0$

This equation holds if and only if $x_1 x_2 - 1 = 0$ or $x_2 - x_1 = 0$.

If $x_2 - x_1 = 0$, then $x_1 = x_2$, which is the condition for being one-one.

However, if $x_1 x_2 - 1 = 0$, i.e., $x_1 x_2 = 1$, then $f(x_1) = f(x_2)$ can be true even if $x_1 \neq x_2$.

For example, let $x_1 = 2$. Then $x_2$ must satisfy $2 x_2 = 1$, so $x_2 = \frac{1}{2}$. Here, $x_1 = 2$ and $x_2 = \frac{1}{2}$ are distinct values in $\mathbb{R}$.

Let's evaluate the function at these points:

$f(2) = \frac{2}{2^2 + 1} = \frac{2}{4 + 1} = \frac{2}{5}$

$f\left(\frac{1}{2}\right) = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2 + 1} = \frac{\frac{1}{2}}{\frac{1}{4} + 1} = \frac{\frac{1}{2}}{\frac{1+4}{4}} = \frac{\frac{1}{2}}{\frac{5}{4}} = \frac{1}{2} \times \frac{4}{5} = \frac{4}{10} = \frac{2}{5}$

Since $f(2) = f\left(\frac{1}{2}\right) = \frac{2}{5}$ but $2 \neq \frac{1}{2}$, the function $f$ is not one-one.

Not Onto:

A function $f : A \to B$ is onto if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$.

The codomain of $f$ is given as $\mathbb{R}$. To show that $f$ is not onto, we need to find a value $y \in \mathbb{R}$ such that there is no $x \in \mathbb{R}$ for which $f(x) = y$.

Let $y \in \mathbb{R}$. We want to check if the equation $f(x) = y$ has a solution for $x \in \mathbb{R}$.

$y = \frac{x}{x^2 + 1}$

Assume $y \neq 0$. Multiply both sides by $(x^2 + 1)$: $y(x^2 + 1) = x$

$yx^2 + y = x$

Rearranging the terms to form a quadratic equation in $x$: $yx^2 - x + y = 0$

For this quadratic equation to have real solutions for $x$, the discriminant must be non-negative.

The discriminant $\Delta$ is given by $\Delta = b^2 - 4ac$, where $a=y$, $b=-1$, and $c=y$.

$\Delta = (-1)^2 - 4(y)(y) = 1 - 4y^2$

For real solutions for $x$ to exist, we must have $\Delta \ge 0$.

$1 - 4y^2 \ge 0$

$1 \ge 4y^2$

$y^2 \le \frac{1}{4}$

This inequality $y^2 \le \frac{1}{4}$ is equivalent to $-\frac{1}{2} \le y \le \frac{1}{2}$.

If $y=0$, the equation is $0 = \frac{x}{x^2+1}$ which gives $x=0$. So $y=0$ is in the range, and $0 \in [-\frac{1}{2}, \frac{1}{2}]$.

This means that the function $f(x)$ can only take values in the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$. This is the range of the function $f$.

The codomain of $f$ is given as $\mathbb{R}$. However, the range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$, which is a proper subset of the codomain $\mathbb{R}$.

For example, let $y = 1$. The value $1$ is in the codomain $\mathbb{R}$. However, $1$ is not in the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$. This means there is no real number $x$ such that $f(x) = 1$.

Since there are elements in the codomain $\mathbb{R}$ (like $y=1$) for which there is no corresponding element in the domain $\mathbb{R}$, the function $f$ is not onto.

Therefore, the function $f(x) = \frac{x}{x^2 + 1}$ is neither one-one nor onto.

Given:

Two functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ defined by:

$f(x) = |x| + x$, for all $x \in \mathbb{R}$

$g(x) = |x| - x$, for all $x \in \mathbb{R}$

To Find:

The composite functions $f \circ g$ and $g \circ f$.

Solution:

Let's first write the functions $f(x)$ and $g(x)$ without the absolute value sign, by considering the cases $x \ge 0$ and $x < 0$.

For $f(x) = |x| + x$:

If $x \ge 0$, then $|x| = x$. So, $f(x) = x + x = 2x$.

If $x < 0$, then $|x| = -x$. So, $f(x) = -x + x = 0$.

Thus, the function $f(x)$ can be written as:

For $g(x) = |x| - x$:

If $x \ge 0$, then $|x| = x$. So, $g(x) = x - x = 0$.

If $x < 0$, then $|x| = -x$. So, $g(x) = -x - x = -2x$.

Thus, the function $g(x)$ can be written as:

Now, let's find the composite function $f \circ g(x) = f(g(x))$.

We need to consider the cases based on the definition of $g(x)$:

Case 1: $x \ge 0$

When $x \ge 0$, $g(x) = 0$.

So, $f(g(x)) = f(0)$.

Since $0 \ge 0$, we use the first case for $f(y)$ (where $y=0$): $f(y) = 2y$.

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

Thus, for $x \ge 0$, $f \circ g(x) = 0$.

Case 2: $x < 0$

When $x < 0$, $g(x) = -2x$.

Since $x < 0$, $-2x$ is a positive number (e.g., if $x=-1$, $-2x=2$). So, $g(x) > 0$.

Since $g(x) > 0$, which satisfies the condition $y \ge 0$, we use the first case for $f(y)$ (where $y=g(x)=-2x$): $f(y) = 2y$.

$f(g(x)) = f(-2x) = 2(-2x) = -4x$.

Thus, for $x < 0$, $f \circ g(x) = -4x$.

Combining the cases, the composite function $f \circ g(x)$ is:

Next, let's find the composite function $g \circ f(x) = g(f(x))$.

We need to consider the cases based on the definition of $f(x)$:

Case 1: $x \ge 0$

When $x \ge 0$, $f(x) = 2x$.

Since $x \ge 0$, $2x$ is a non-negative number (e.g., if $x=1$, $2x=2$; if $x=0$, $2x=0$). So, $f(x) \ge 0$.

Since $f(x) \ge 0$, we use the first case for $g(y)$ (where $y=f(x)=2x$): $g(y) = 0$.

$g(f(x)) = g(2x) = 0$.

Thus, for $x \ge 0$, $g \circ f(x) = 0$.

Case 2: $x < 0$

When $x < 0$, $f(x) = 0$.

Since $0 \ge 0$, we use the first case for $g(y)$ (where $y=0$): $g(y) = 0$.

$g(f(x)) = g(0) = 0$.

Thus, for $x < 0$, $g \circ f(x) = 0$.

Combining the cases, the composite function $g \circ f(x)$ is:

Summary of Results:

$f \circ g(x) = \begin{cases} 0 & , & x \ge 0 \\ -4x & , & x < 0 \end{cases}$

$g \circ f(x) = 0$

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 4x + 5$ for all $x \in \mathbb{R}$.

To Show:

The function $f$ is invertible and to find its inverse $f^{-1}$.

Proof of Invertibility and Finding Inverse:

A function is invertible if and only if it is a bijection (both one-one and onto).

One-one (Injectivity):

Let $x_1, x_2 \in \mathbb{R}$ be such that $f(x_1) = f(x_2)$.

$4x_1 + 5 = 4x_2 + 5$

Subtracting 5 from both sides, we get:

$4x_1 = 4x_2$

Dividing both sides by 4, we get:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-one.

Onto (Surjectivity):

Let $y$ be an arbitrary element in the codomain $\mathbb{R}$. We need to find an element $x$ in the domain $\mathbb{R}$ such that $f(x) = y$.

Set $f(x) = y$:

$4x + 5 = y$

Solving for $x$ in terms of $y$:

$4x = y - 5$

$x = \frac{y - 5}{4}$

Since $y \in \mathbb{R}$, $\frac{y - 5}{4}$ is always a real number. Thus, for every $y$ in the codomain $\mathbb{R}$, there exists an $x = \frac{y-5}{4}$ in the domain $\mathbb{R}$ such that $f(x) = y$.

This shows that for every element in the codomain, there is a preimage in the domain. Therefore, the function $f$ is onto.

Since the function $f$ is both one-one and onto, it is a bijection. Hence, $f$ is invertible.

Finding the Inverse Function ($f^{-1}$):

To find the inverse function $f^{-1}(x)$, we let $y = f(x)$ and solve for $x$ in terms of $y$. Then, we swap $x$ and $y$.

We already did the solving part when checking for surjectivity:

$y = 4x + 5$

$x = \frac{y - 5}{4}$

So, the inverse function can be written as $f^{-1}(y) = \frac{y - 5}{4}$.

Replacing the variable $y$ with $x$ (as is conventional for the argument of the inverse function), we get:

$f^{-1}(x) = \frac{x - 5}{4}$

The inverse function $f^{-1}$ maps from $\mathbb{R}$ to $\mathbb{R}$.

Conclusion:

The function $f(x) = 4x + 5$ is invertible, and its inverse is $f^{-1}(x) = \frac{x - 5}{4}$.

Given:

The set of rational numbers $\mathbb{Q}$.

Four binary operations defined on $\mathbb{Q}$ as follows:

(i) $a * b = a - b$

(ii) $a * b = \frac{ab}{4}$

(iii) $a * b = a - b + ab$

(iv) $a * b = ab^2$

To Find:

Which of the given binary operations are associative.

Solution:

A binary operation $*$ on a set $\mathbb{Q}$ is said to be associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{Q}$.

We will check each operation for associativity:

(i) $a * b = a - b$

Let $a, b, c \in \mathbb{Q}$.

Left Hand Side (LHS): $(a * b) * c = (a - b) * c = (a - b) - c = a - b - c$.

Right Hand Side (RHS): $a * (b * c) = a * (b - c) = a - (b - c) = a - b + c$.

For the operation to be associative, we need $a - b - c = a - b + c$ for all $a, b, c \in \mathbb{Q}$.

This implies $-c = c$, or $2c = 0$, which means $c = 0$. This is not true for all $c \in \mathbb{Q}$.

For example, let $a=1, b=2, c=3$.

$(1 * 2) * 3 = (1 - 2) * 3 = (-1) * 3 = -1 - 3 = -4$.

$1 * (2 * 3) = 1 * (2 - 3) = 1 * (-1) = 1 - (-1) = 1 + 1 = 2$.

Since $-4 \neq 2$, the operation $a * b = a - b$ is not associative.

(ii) $a * b = \frac{ab}{4}$

Let $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = \left(\frac{ab}{4}\right) * c = \frac{\left(\frac{ab}{4}\right)c}{4} = \frac{abc}{16}$.

RHS: $a * (b * c) = a * \left(\frac{bc}{4}\right) = \frac{a\left(\frac{bc}{4}\right)}{4} = \frac{abc}{16}$.

Since $(a * b) * c = \frac{abc}{16}$ and $a * (b * c) = \frac{abc}{16}$ for all $a, b, c \in \mathbb{Q}$, the operation $a * b = \frac{ab}{4}$ is associative.

(iii) $a * b = a - b + ab$

Let $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = (a - b + ab) * c = (a - b + ab) - c + (a - b + ab)c = a - b + ab - c + ac - bc + abc$.

RHS: $a * (b * c) = a * (b - c + bc) = a - (b - c + bc) + a(b - c + bc) = a - b + c - bc + ab - ac + abc$.

For the operation to be associative, we need $a - b + ab - c + ac - bc + abc = a - b + c - bc + ab - ac + abc$ for all $a, b, c \in \mathbb{Q}$.

Simplifying the equation, we need $-c + ac = c - ac$ for all $a, c \in \mathbb{Q}$.

This simplifies to $2ac - 2c = 0$, or $2c(a - 1) = 0$. This implies $c=0$ or $a=1$. This is not true for all $a, c \in \mathbb{Q}$.

For example, let $a=2, b=3, c=4$.

$(2 * 3) * 4 = (2 - 3 + 2 \times 3) * 4 = (-1 + 6) * 4 = 5 * 4 = 5 - 4 + 5 \times 4 = 1 + 20 = 21$.

$2 * (3 * 4) = 2 * (3 - 4 + 3 \times 4) = 2 * (-1 + 12) = 2 * 11 = 2 - 11 + 2 \times 11 = -9 + 22 = 13$.

Since $21 \neq 13$, the operation $a * b = a - b + ab$ is not associative.

(iv) $a * b = ab^2$

Let $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = (ab^2) * c = (ab^2)c^2 = ab^2c^2$.

RHS: $a * (b * c) = a * (bc^2) = a(bc^2)^2 = a(b^2c^4) = ab^2c^4$.

For the operation to be associative, we need $ab^2c^2 = ab^2c^4$ for all $a, b, c \in \mathbb{Q}$.

This implies $ab^2c^2 - ab^2c^4 = 0$, or $ab^2c^2(1 - c^2) = 0$. This implies $a=0$ or $b=0$ or $c=0$ or $c=1$ or $c=-1$. This is not true for all $a, b, c \in \mathbb{Q}$.

For example, let $a=1, b=1, c=2$.

$(1 * 1) * 2 = (1 \times 1^2) * 2 = 1 * 2 = 1 \times 2^2 = 1 \times 4 = 4$.

$1 * (1 * 2) = 1 * (1 \times 2^2) = 1 * (1 \times 4) = 1 * 4 = 1 \times 4^2 = 1 \times 16 = 16$.

Since $4 \neq 16$, the operation $a * b = ab^2$ is not associative.

Conclusion:

Out of the given binary operations, only operation (ii) is associative.

Given:

Set $N = \{1, 2, 3, ...\}$ (the set of natural numbers).

Relation $R$ on $N$ defined by $nRm$ if $n$ divides $m$, for $n, m \in N$.

To Find:

Properties of the relation $R$.

Solution:

We examine the properties of the relation $R$ on the set of natural numbers $N$, where $nRm$ means that $n$ divides $m$ (denoted as $n | m$).

Reflexivity:

A relation $R$ on a set $A$ is reflexive if $aRa$ for every $a \in A$.

For any natural number $n \in N$, does $nRn$? This means, does $n$ divide $n$?

Yes, any natural number $n$ divides itself ($n = 1 \times n$).

So, $nRn$ for all $n \in N$. The relation $R$ is reflexive.

Symmetry:

A relation $R$ on a set $A$ is symmetric if whenever $aRb$, then $bRa$ for all $a, b \in A$.

For any natural numbers $n, m \in N$, if $nRm$ (i.e., $n$ divides $m$), does $mRn$ (i.e., $m$ divides $n$)?

If $n$ divides $m$, there exists an integer $k$ such that $m = kn$.

If $m$ divides $n$, there exists an integer $l$ such that $n = lm$.

If both $n$ divides $m$ and $m$ divides $n$, then $m = kn$ and $n = lm$. Substituting $n$ in the first equation, $m = k(lm) = (kl)m$.

Since $m \in N$, $m \ge 1$, so we can divide by $m$ to get $1 = kl$.

Since $k$ and $l$ are integers and $n, m \in N$, $k$ and $l$ must be positive integers. The only positive integer solutions to $kl=1$ are $k=1$ and $l=1$.

This implies $m = 1 \times n = n$ and $n = 1 \times m = m$. So, $n=m$.

This means $nRm \implies mRn$ only if $n=m$. If $n \neq m$ and $nRm$, then $m$ does not divide $n$.

For example, let $n=2$ and $m=4$. $2 \in N$ and $4 \in N$.

$2R4$ is true because 2 divides 4.

Is $4R2$ true? This means does 4 divide 2?

No, 4 does not divide 2.

Since we found a pair $(2, 4)$ such that $2R4$ is true but $4R2$ is false, the relation $R$ is not symmetric.

Transitivity:

A relation $R$ on a set $A$ is transitive if whenever $aRb$ and $bRc$, then $aRc$ for all $a, b, c \in A$.

For any natural numbers $n, m, p \in N$, if $nRm$ (i.e., $n$ divides $m$) and $mRp$ (i.e., $m$ divides $p$), does $nRp$ (i.e., $n$ divides $p$)?

If $n$ divides $m$, there exists an integer $k$ such that $m = kn$. Since $n, m \in N$, $k$ must be a positive integer ($k \in N$).

If $m$ divides $p$, there exists an integer $l$ such that $p = lm$. Since $m, p \in N$, $l$ must be a positive integer ($l \in N$).

Substitute the expression for $m$ from the first equation into the second equation:

$p = l(kn)$

$p = (lk)n$

Since $l$ and $k$ are positive integers, $lk$ is also a positive integer.

This means $p$ is an integer multiple of $n$. Therefore, $n$ divides $p$.

So, if $nRm$ and $mRp$, then $nRp$. The relation $R$ is transitive.

In summary, the relation $R$ is reflexive and transitive, but not symmetric.

Comparing this with the given options, option (D) matches our findings.

The correct answer is (D) Reflexive, transitive but not symmetric.

Given:

Set $L$: the set of all straight lines in a plane.

Relation $R$ on $L$ defined by $lRm$ if and only if line $l$ is perpendicular to line $m$ ($l \perp m$), for all $l, m \in L$.

To Find:

Which property (reflexive, symmetric, transitive) the relation $R$ possesses.

Solution:

We examine the properties of the relation $R$ on the set of lines $L$, where $lRm$ means $l \perp m$.

Reflexivity:

A relation $R$ on a set $A$ is reflexive if $aRa$ for every $a \in A$.

For any line $l \in L$, does $lRl$? This means, is line $l$ perpendicular to itself ($l \perp l$)?

A line cannot be perpendicular to itself. The angle a line makes with itself is $0^\circ$ or $180^\circ$, not $90^\circ$.

So, $lRl$ is false for all $l \in L$. The relation $R$ is not reflexive.

Symmetry:

A relation $R$ on a set $A$ is symmetric if whenever $aRb$, then $bRa$ for all $a, b \in A$.

For any lines $l, m \in L$, if $lRm$ (i.e., $l \perp m$), does $mRl$ (i.e., $m \perp l$)?

If line $l$ is perpendicular to line $m$, then the angle between them is $90^\circ$. This implies that line $m$ is also perpendicular to line $l$.

So, if $lRm$, then $mRl$ is always true.

The relation $R$ is symmetric.

Transitivity:

A relation $R$ on a set $A$ is transitive if whenever $aRb$ and $bRc$, then $aRc$ for all $a, b, c \in A$.

For any lines $l, m, p \in L$, if $lRm$ (i.e., $l \perp m$) and $mRp$ (i.e., $m \perp p$), does $lRp$ (i.e., $l \perp p$)?

If line $l$ is perpendicular to line $m$, and line $m$ is perpendicular to line $p$, then line $l$ must be parallel to line $p$.

For example, let $l$ be the x-axis. Let $m$ be the y-axis. Then $l \perp m$.

Let $p$ be the line $y=5$. Then $m \perp p$ (y-axis is perpendicular to the horizontal line $y=5$).

Is $l \perp p$? (Is the x-axis perpendicular to the line $y=5$?).

No, the x-axis ($y=0$) is parallel to the line $y=5$.

Since we found lines $l, m, p$ such that $lRm$ and $mRp$ are true, but $lRp$ is false, the relation $R$ is not transitive.

In summary, the relation $R$ is not reflexive, is symmetric, and is not transitive.

Comparing this with the given options:

(A) Reflexive - False

(B) Symmetric - True

(C) Transitive - False

(D) None of these - Since the relation is symmetric, option (B) describes a property it possesses.

The correct answer is (B) symmetric.

Given:

Set $N = \{1, 2, 3, ...\}$ (the set of natural numbers).

Function $f : N \to N$ defined by $f(n) = 2n + 3$ for all $n \in N$.

To Find:

Whether the function $f$ is surjective, injective, bijective, or none of these.

Solution:

We examine the properties of the function $f(n) = 2n + 3$ defined from the set of natural numbers to the set of natural numbers.

Injective (One-one):

A function $f : A \to B$ is injective if for any $a_1, a_2 \in A$, $f(a_1) = f(a_2)$ implies $a_1 = a_2$.

Let $n_1, n_2 \in N$ be such that $f(n_1) = f(n_2)$.

$2n_1 + 3 = 2n_2 + 3$

Subtracting 3 from both sides:

$2n_1 = 2n_2$

Dividing both sides by 2:

$n_1 = n_2$

Since $f(n_1) = f(n_2)$ implies $n_1 = n_2$, the function $f$ is injective (one-one).

Surjective (Onto):

A function $f : A \to B$ is surjective if for every element $b$ in the codomain $B$, there exists at least one element $a$ in the domain $A$ such that $f(a) = b.

The codomain of $f$ is $N = \{1, 2, 3, ...\}$.

Let $y$ be an element in the codomain $N$. We need to check if there exists an $n \in N$ such that $f(n) = y$.

$2n + 3 = y$

Solving for $n$:

$2n = y - 3$

$n = \frac{y - 3}{2}$

For $n$ to be in the domain $N$, two conditions must be met:

1. $n$ must be a natural number (i.e., $n \in \{1, 2, 3, ...\}$).

2. $n$ must be an integer and $n \ge 1$.

For $n = \frac{y-3}{2}$ to be an integer, $y-3$ must be an even number, which means $y$ must be an odd integer.

Also, for $n \ge 1$, we need $\frac{y-3}{2} \ge 1$, which implies $y-3 \ge 2$, or $y \ge 5$.

So, for a value $y$ in the codomain $N$ to have a preimage in the domain $N$, $y$ must be an odd natural number greater than or equal to 5.

Consider elements in the codomain $N$ that do not satisfy these conditions.

For example, let $y = 1$. $1 \in N$. Does there exist $n \in N$ such that $f(n) = 1$?

$2n + 3 = 1 \implies 2n = -2 \implies n = -1$.

Since $-1$ is not a natural number ($-1 \notin N$), there is no $n \in N$ such that $f(n) = 1$.

Similarly, for $y=2$, $2n+3 = 2 \implies 2n = -1 \implies n = -\frac{1}{2}$, which is not in $N$.

For $y=3$, $2n+3 = 3 \implies 2n = 0 \implies n = 0$, which is not in $N$.

For $y=4$, $2n+3 = 4 \implies 2n = 1 \implies n = \frac{1}{2}$, which is not in $N$.

For $y=5$, $2n+3 = 5 \implies 2n = 2 \implies n = 1$. Here, $1 \in N$, and $f(1) = 2(1)+3 = 5$. So 5 has a preimage.

For $y=6$, $2n+3 = 6 \implies 2n = 3 \implies n = \frac{3}{2}$, which is not in $N$.

For $y=7$, $2n+3 = 7 \implies 2n = 4 \implies n = 2$. Here, $2 \in N$, and $f(2) = 2(2)+3 = 7$. So 7 has a preimage.

The range of the function $f$ is $\{5, 7, 9, 11, ...\}$, which is the set of odd natural numbers greater than or equal to 5.

The codomain of $f$ is $N = \{1, 2, 3, 4, 5, 6, 7, 8, 9, ...\}$.

Since the range $\{5, 7, 9, ...\}$ is a proper subset of the codomain $\{1, 2, 3, ...\}$, the function $f$ is not surjective (not onto).

Since $f$ is injective but not surjective, it is not bijective.

Comparing with the options, option (B) states that $f$ is injective, which is true.

The correct answer is (B) injective.

Given:

Set A with $|A| = 3$ elements.

Set B with $|B| = 4$ elements.

To Find:

The number of injective mappings (one-one functions) that can be defined from set A to set B.

Solution:

Let the elements of set A be $a_1, a_2, a_3$.

Let the elements of set B be $b_1, b_2, b_3, b_4$.

A function $f: A \to B$ is injective if distinct elements in A are mapped to distinct elements in B.

We need to define the mapping for each element in A such that no two elements in A are mapped to the same element in B.

Consider the mapping of the first element $a_1 \in A$.

The element $a_1$ can be mapped to any of the 4 elements in set B.

Number of choices for $f(a_1)$ is 4.

Now, consider the mapping of the second element $a_2 \in A$.

For the function to be injective, $a_2$ must be mapped to an element in B that is different from the element $a_1$ is mapped to.

Since one element in B has already been taken by $f(a_1)$, there are $4 - 1 = 3$ remaining elements in B that $a_2$ can be mapped to.

Number of choices for $f(a_2)$ is 3.

Finally, consider the mapping of the third element $a_3 \in A$.

For the function to be injective, $a_3$ must be mapped to an element in B that is different from the elements that $a_1$ and $a_2$ are mapped to.

Since two distinct elements in B have already been taken by $f(a_1)$ and $f(a_2)$, there are $4 - 2 = 2$ remaining elements in B that $a_3$ can be mapped to.

Number of choices for $f(a_3)$ is 2.

The total number of ways to define such a mapping is the product of the number of choices for each element in A.

Total number of injective mappings = (Choices for $f(a_1)$) $\times$ (Choices for $f(a_2)$) $\times$ (Choices for $f(a_3)$)

Total number of injective mappings = $4 \times 3 \times 2 = 24$.

In general, if $|A| = m$ and $|B| = n$, the number of injective mappings from A to B is given by the number of permutations of $n$ items taken $m$ at a time, denoted as $P(n, m)$ or $_nP_m$, provided $m \le n$.

$P(n, m) = \frac{n!}{(n - m)!}$

In this case, $|A| = m = 3$ and $|B| = n = 4$. Since $3 \le 4$, injective mappings are possible.

Number of injective mappings = $P(4, 3) = \frac{4!}{(4 - 3)!} = \frac{4!}{1!} = \frac{4 \times 3 \times 2 \times 1}{1} = 24$.

Comparing with the given options, option (C) is 24.

The correct answer is (C) 24.

Given:

Function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \sin x$.

Function $g : \mathbb{R} \to \mathbb{R}$ defined by $g(x) = x^2$.

To Find:

The composite function $f \circ g$.

Solution:

The composite function $f \circ g$ is defined as $f \circ g(x) = f(g(x))$.

First, we take the input $x$ and apply the function $g$ to it.

$g(x) = x^2$

Next, we take the output of $g(x)$ and apply the function $f$ to it.

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

The function $f$ takes an input and returns the sine of that input. So, $f(x^2)$ means the sine of $x^2$.

$f(x^2) = \sin(x^2)$

The notation $\sin x^2$ is commonly used to represent $\sin(x^2)$.

Therefore, $f \circ g(x) = \sin x^2$.

Comparing this with the given options:

(A) $x^2 \sin x$ is the product of $x^2$ and $\sin x$.

(B) $(\sin x)^2$ is the square of $\sin x$, which is also written as $\sin^2 x$. This is the composition $g(f(x)) = (\sin x)^2$.

(C) $\sin x^2$ is the sine of $x^2$.

(D) $\frac{\sin x}{x^2}$ is the quotient of $\sin x$ and $x^2$.

Option (C) matches our result.

The correct answer is (C) sin x².

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3x - 4$ for all $x \in \mathbb{R}$.

To Find:

The inverse function $f^{-1}(x)$.

Solution:

To find the inverse function $f^{-1}(x)$, we follow these steps:

1. Let $y = f(x)$.

2. Solve the equation for $x$ in terms of $y$.

3. Replace $y$ with $x$ to get the expression for $f^{-1}(x)$.

Step 1: Let $y = f(x)$.

$y = 3x - 4$

Step 2: Solve for $x$ in terms of $y$.

Add 4 to both sides:

$y + 4 = 3x$

Divide both sides by 3:

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

Step 3: Replace $y$ with $x$ to get $f^{-1}(x)$.

$f^{-1}(x) = \frac{x + 4}{3}$

We can verify this by checking if $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.

$f(f^{-1}(x)) = f\left(\frac{x+4}{3}\right) = 3\left(\frac{x+4}{3}\right) - 4 = (x+4) - 4 = x$.

$f^{-1}(f(x)) = f^{-1}(3x - 4) = \frac{(3x - 4) + 4}{3} = \frac{3x}{3} = x$.

Since both compositions result in $x$, the inverse function is correct.

Comparing our result $f^{-1}(x) = \frac{x + 4}{3}$ with the given options:

(A) $\frac{x + 4}{3}$ matches our result.

(B) $\frac{x}{3} - 4 = \frac{x - 12}{3}$, which is incorrect.

(C) $3x + 4$, which is incorrect.

(D) None of these - This is incorrect since option (A) is the correct inverse.

The correct answer is (A) $\frac{x + 4}{3}$.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 + 1$ for all $x \in \mathbb{R}$.

To Find:

The pre-images of 17 and -3.

Solution:

The pre-image of a value $y$ in the codomain is the set of all elements $x$ in the domain such that $f(x) = y$.

Pre-image of $y = \{x \in \mathbb{R} \mid f(x) = y\}$.

Pre-image of 17:

We need to find the set of $x \in \mathbb{R}$ such that $f(x) = 17$.

$x^2 + 1 = 17$

$x^2 = 17 - 1$

$x^2 = 16$

Taking the square root of both sides:

$x = \pm \sqrt{16}$

$x = \pm 4$

So, the values of $x$ are $4$ and $-4$. Both $4 \in \mathbb{R}$ and $-4 \in \mathbb{R}$.

The pre-image of 17 is the set $\{4, -4\}$.

Pre-image of -3:

We need to find the set of $x \in \mathbb{R}$ such that $f(x) = -3$.

$x^2 + 1 = -3$

$x^2 = -3 - 1$

$x^2 = -4$

We need to find real numbers $x$ whose square is $-4$. The square of any real number is always non-negative ($x^2 \ge 0$). There is no real number $x$ such that $x^2 = -4$.

Therefore, the pre-image of -3 in $\mathbb{R}$ is the empty set, denoted by $\phi$.

The pre-images of 17 and -3, respectively, are $\{4, -4\}$ and $\phi$.

Comparing this with the given options:

(A) $\phi$, $\{4, – 4\}$ - Incorrect order.

(B) $\{3, – 3\}$, $\phi$ - Incorrect pre-image for 17.

(C) $\{4, –4\}$, $\phi$ - Matches our result.

(D) $\{4, – 4\}$, $\{2, – 2\}$ - Incorrect pre-image for -3.

The correct answer is (C) {4, –4}, $\phi$.

Given:

Set $\mathbb{R}$: the set of real numbers.

Relation $R$ on $\mathbb{R}$ defined by $xRy$ if and only if $x - y + \sqrt{2}$ is an irrational number, for all $x, y \in \mathbb{R}$.

To Find:

Which property (reflexive, symmetric, transitive) the relation $R$ possesses.

Solution:

We examine the properties of the relation $R$ on the set of real numbers $\mathbb{R}$, where $xRy$ means $x - y + \sqrt{2} \in \mathbb{R} \setminus \mathbb{Q}$.

Reflexivity:

A relation $R$ on a set $A$ is reflexive if $aRa$ for every $a \in A$.

For any real number $x \in \mathbb{R}$, does $xRx$? This means, is $x - x + \sqrt{2}$ an irrational number?

$x - x + \sqrt{2} = 0 + \sqrt{2} = \sqrt{2}$

Since $\sqrt{2}$ is an irrational number, the condition is true for all $x \in \mathbb{R}$.

So, $xRx$ for all $x \in \mathbb{R}$. The relation $R$ is reflexive.

Symmetry:

A relation $R$ on a set $A$ is symmetric if whenever $aRb$, then $bRa$ for all $a, b \in A$.

For any real numbers $x, y \in \mathbb{R}$, if $xRy$ (i.e., $x - y + \sqrt{2}$ is irrational), does $yRx$ (i.e., $y - x + \sqrt{2}$ is irrational)?

Let $xRy$ be true. This means $x - y + \sqrt{2} = I_1$, where $I_1$ is an irrational number.

Now consider $y - x + \sqrt{2}$.

$y - x + \sqrt{2} = -(x - y) + \sqrt{2} = -((x - y + \sqrt{2}) - \sqrt{2}) + \sqrt{2} = -(I_1 - \sqrt{2}) + \sqrt{2} = -I_1 + \sqrt{2} + \sqrt{2} = -I_1 + 2\sqrt{2}$.

If $I_1$ is irrational, is $-I_1 + 2\sqrt{2}$ always irrational?

Consider a counterexample. Let $x = 1 + \sqrt{2}$ and $y = 1$. Both are real numbers.

$x - y + \sqrt{2} = (1 + \sqrt{2}) - 1 + \sqrt{2} = 2\sqrt{2}$. This is irrational. So, $xRy$ is true for $x = 1 + \sqrt{2}, y = 1$.

Now check $yRx$ for $x = 1 + \sqrt{2}, y = 1$.

$y - x + \sqrt{2} = 1 - (1 + \sqrt{2}) + \sqrt{2} = 1 - 1 - \sqrt{2} + \sqrt{2} = 0$. This is a rational number.

So, $yRx$ is false for $x = 1 + \sqrt{2}, y = 1$.

Since we found a pair $(x, y)$ such that $xRy$ is true but $yRx$ is false, the relation $R$ is not symmetric.

Transitivity:

A relation $R$ on a set $A$ is transitive if whenever $aRb$ and $bRc$, then $aRc$ for all $a, b, c \in A.

For any real numbers $x, y, z \in \mathbb{R}$, if $xRy$ (i.e., $x - y + \sqrt{2}$ is irrational) and $yRz$ (i.e., $y - z + \sqrt{2}$ is irrational), does $xRz$ (i.e., $x - z + \sqrt{2}$ is irrational)?

Let $xRy$ be true, so $x - y + \sqrt{2} = I_1$, where $I_1$ is an irrational number.

Let $yRz$ be true, so $y - z + \sqrt{2} = I_2$, where $I_2$ is an irrational number.

We want to check if $x - z + \sqrt{2}$ is always irrational.

We know that $(x - y) + (y - z) = x - z$.

From the given conditions:

$x - y = I_1 - \sqrt{2}$

$y - z = I_2 - \sqrt{2}$

Adding these equations:

$(x - y) + (y - z) = (I_1 - \sqrt{2}) + (I_2 - \sqrt{2})$

$x - z = I_1 + I_2 - 2\sqrt{2}$

Now, consider $x - z + \sqrt{2}$:

$x - z + \sqrt{2} = (I_1 + I_2 - 2\sqrt{2}) + \sqrt{2} = I_1 + I_2 - \sqrt{2}$

If $I_1$ and $I_2$ are irrational numbers, is $I_1 + I_2 - \sqrt{2}$ always irrational?

Consider a counterexample. Let $I_1 = \frac{\sqrt{2}}{2}$ and $I_2 = \frac{\sqrt{2}}{2}$. Both are irrational numbers.

Then $I_1 + I_2 - \sqrt{2} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - \sqrt{2} = \sqrt{2} - \sqrt{2} = 0$. This is a rational number.

Can we find $x, y, z \in \mathbb{R}$ such that $x - y + \sqrt{2} = \frac{\sqrt{2}}{2}$ and $y - z + \sqrt{2} = \frac{\sqrt{2}}{2}$?

$x - y = \frac{\sqrt{2}}{2} - \sqrt{2} = -\frac{\sqrt{2}}{2}$

$y - z = \frac{\sqrt{2}}{2} - \sqrt{2} = -\frac{\sqrt{2}}{2}$

Let $y = 0$. Then $x = -\frac{\sqrt{2}}{2}$ and $z = y - (-\frac{\sqrt{2}}{2}) = 0 + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$.

So, let $x = -\frac{\sqrt{2}}{2}, y = 0, z = \frac{\sqrt{2}}{2}$. All are real numbers.

Check $xRy$: $x - y + \sqrt{2} = -\frac{\sqrt{2}}{2} - 0 + \sqrt{2} = \frac{\sqrt{2}}{2}$. Irrational. $xRy$ is true.

Check $yRz$: $y - z + \sqrt{2} = 0 - \frac{\sqrt{2}}{2} + \sqrt{2} = \frac{\sqrt{2}}{2}$. Irrational. $yRz$ is true.

Check $xRz$: $x - z + \sqrt{2} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} + \sqrt{2} = -\sqrt{2} + \sqrt{2} = 0$. Rational. $xRz$ is false.

Since we found $x, y, z$ such that $xRy$ and $yRz$ are true but $xRz$ is false, the relation $R$ is not transitive.

In summary, the relation $R$ is reflexive, but not symmetric and not transitive.

Comparing this with the given options:

(A) reflexive - This property holds.

(B) symmetric - This property does not hold.

(C) transitive - This property does not hold.

(D) none of these - This is false because the relation is reflexive.

Since the relation is reflexive, option (A) is the correct answer among the choices.

The correct answer is (A) reflexive.

Given:

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

$R$ is the smallest equivalence relation on $A$.

To Find:

The relation $R$.

Solution:

An equivalence relation must satisfy three properties:

1. Reflexivity: For all $a \in A$, $aRa$.

2. Symmetry: For all $a, b \in A$, if $aRb$, then $bRa$.

3. Transitivity: For all $a, b, c \in A$, if $aRb$ and $bRc$, then $aRc$.

We are looking for the "smallest" equivalence relation on A. The smallest relation on A is the empty set $\phi$, but this is generally not reflexive (unless A is empty).

For a relation to be reflexive on $A = \{1, 2, 3\}$, it must contain the pairs $(1, 1)$, $(2, 2)$, and $(3, 3)$.

Let's consider the relation $R_0 = \{(1, 1), (2, 2), (3, 3)\}$.

Check the properties for $R_0$:

1. Reflexivity: Contains $(1, 1), (2, 2), (3, 3)$. Yes, $R_0$ is reflexive.

2. Symmetry: If $(a, b) \in R_0$, then $a=b$. The only pairs are $(1, 1), (2, 2), (3, 3)$. If $(1, 1) \in R_0$, is $(1, 1) \in R_0$? Yes. This holds for all pairs in $R_0$. Yes, $R_0$ is symmetric.

3. Transitivity: If $(a, b) \in R_0$ and $(b, c) \in R_0$, then $(a, c) \in R_0$. If $(a, b) \in R_0$, then $a=b$. If $(b, c) \in R_0$, then $b=c$. So, if $(a, b) \in R_0$ and $(b, c) \in R_0$, we have $a=b=c$. Then $(a, c) = (a, a)$, which is in $R_0$. Yes, $R_0$ is transitive.

Since $R_0$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Any equivalence relation on A must contain all reflexive pairs $\{(1, 1), (2, 2), (3, 3)\}$. $R_0$ contains exactly these pairs.

If we add any other pair $(a, b)$ where $a \neq b$ to $R_0$, say $(1, 2)$, then for symmetry, we must also add $(2, 1)$.

If we have $(1, 2)$ and $(2, 1)$, then for transitivity, we must have $(1, 1)$ [from $(1, 2)$ and $(2, 1)$], $(2, 2)$ [from $(2, 1)$ and $(1, 2)$]. These are already in $R_0$.

Consider the possible equivalence relations on A. An equivalence relation partitions the set A into disjoint equivalence classes. The smallest equivalence relation corresponds to the finest partition, where each element is in its own class.

Partition: $\{1\}, \{2\}, \{3\}$.

The equivalence classes are $[1] = \{1\}$, $[2] = \{2\}$, $[3] = \{3\}$.

The relation R is the union of the Cartesian products of the equivalence classes: $[1] \times [1] \cup [2] \times [2] \cup [3] \times [3]$

$R = (\{1\} \times \{1\}) \cup (\{2\} \times \{2\}) \cup (\{3\} \times \{3\})$

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

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

This is indeed the smallest set of pairs that satisfies the three properties of an equivalence relation on A.

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

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \sqrt{x^2 - 3x + 2}$.

To Find:

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

Solution:

The function $f(x)$ is defined by a square root expression, $\sqrt{x^2 - 3x + 2}$.

For the square root of a real number to be a real number, the expression inside the square root must be non-negative.

So, we require $x^2 - 3x + 2 \ge 0$.

This is a quadratic inequality. Let's find the roots of the quadratic equation $x^2 - 3x + 2 = 0$.

We can factor the quadratic expression:

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

$x(x - 1) - 2(x - 1) = 0$

$(x - 1)(x - 2) = 0$

The roots are $x - 1 = 0 \implies x = 1$ and $x - 2 = 0 \implies x = 2$.

These roots divide the real number line into three intervals: $(-\infty, 1)$, $(1, 2)$, and $(2, \infty)$.

The quadratic $x^2 - 3x + 2$ is a parabola opening upwards (since the coefficient of $x^2$ is positive, which is 1).

The parabola is above or on the x-axis when $x \le 1$ or $x \ge 2$.

We can also test a value in each interval:

- For $x < 1$, let $x=0$. $0^2 - 3(0) + 2 = 2$. $2 \ge 0$. The inequality holds in $(-\infty, 1]$.

- For $1 < x < 2$, let $x=1.5$. $(1.5)^2 - 3(1.5) + 2 = 2.25 - 4.5 + 2 = -0.25$. $-0.25 < 0$. The inequality does not hold in $(1, 2)$.

- For $x > 2$, let $x=3$. $3^2 - 3(3) + 2 = 9 - 9 + 2 = 2$. $2 \ge 0$. The inequality holds in $[2, \infty)$.

The inequality $x^2 - 3x + 2 \ge 0$ is satisfied when $x \le 1$ or $x \ge 2$.

In interval notation, this is $(-\infty, 1] \cup [2, \infty)$.

The domain of the function $f(x)$ is the set of all real numbers $x$ for which the expression under the square root is non-negative.

Domain of $f = \{x \in \mathbb{R} \mid x^2 - 3x + 2 \ge 0\}$.

Domain of $f = (-\infty, 1] \cup [2, \infty)$.

The domain of the function is $(-\infty, 1] \cup [2, \infty)$.

Given:

Set A with $|A| = n$ elements.

We are considering functions $f: A \to A$.

We are interested in injective functions from A onto itself.

To Find:

The number of injective functions from A to A.

Solution:

Let the elements of set A be $a_1, a_2, ..., a_n$. The domain and codomain are the same set A.

An injective function (one-one function) $f: A \to A$ maps distinct elements in A to distinct elements in A.

Let's consider the mapping of each element in A.

The first element $a_1 \in A$ can be mapped to any of the $n$ elements in the codomain A.

Number of choices for $f(a_1)$ is $n$.

The second element $a_2 \in A$ must be mapped to an element in the codomain A that is different from where $a_1$ was mapped (since the function must be injective).

Number of choices for $f(a_2)$ is $n - 1$.

The third element $a_3 \in A$ must be mapped to an element in the codomain A that is different from where $a_1$ and $a_2$ were mapped.

Number of choices for $f(a_3)$ is $n - 2$.

...and so on.

The $k$-th element $a_k \in A$ must be mapped to an element in the codomain A that is different from where the first $k-1$ elements were mapped.

Number of choices for $f(a_k)$ is $n - (k - 1) = n - k + 1$.

Finally, the last element $a_n \in A$ must be mapped to the one remaining element in the codomain A that has not been used as an image for the previous $n-1$ elements.

Number of choices for $f(a_n)$ is $n - (n - 1) = 1$.

The total number of injective functions from A to A is the product of the number of choices for each element:

Total number of injective functions = $n \times (n - 1) \times (n - 2) \times ... \times 1$.

This product is the definition of the factorial of $n$, denoted by $n!$.

Number of injective functions = $n!$.

When the domain and codomain have the same finite number of elements, a function from the set to itself is injective if and only if it is surjective. An injective and surjective function is a bijection. A bijection from a set to itself is also called a permutation of the set.

The number of permutations of a set with $n$ elements is $n!$.

The total number of injective functions from A onto itself is $n!$.

Given:

Set $\mathbb{Z}$: the set of integers.

Relation $R$ on $\mathbb{Z}$ defined by $aRb$ if $a - b$ is divisible by 3, for all $a, b \in \mathbb{Z}$.

To Find:

The number of pairwise disjoint subsets (equivalence classes) into which the relation R partitions the set $\mathbb{Z}$.

Solution:

The relation $aRb$ means $a - b$ is divisible by 3, which can be written as $a - b \equiv 0 \pmod 3$, or $a \equiv b \pmod 3$.

This is the congruence modulo 3 relation on $\mathbb{Z}$.

It is a standard result that the congruence modulo $n$ relation on $\mathbb{Z}$ is an equivalence relation.

Let's quickly verify the properties:

1. Reflexivity: For any $a \in \mathbb{Z}$, $a - a = 0$. $0$ is divisible by 3. So $aRa$. Reflexive.

2. Symmetry: If $aRb$, then $a - b$ is divisible by 3. So $a - b = 3k$ for some integer $k$. Then $b - a = -(a - b) = -3k = 3(-k)$. Since $-k$ is also an integer, $b - a$ is divisible by 3. So $bRa$. Symmetric.

3. Transitivity: If $aRb$ and $bRc$, then $a - b$ is divisible by 3 and $b - c$ is divisible by 3. So $a - b = 3k$ and $b - c = 3l$ for some integers $k, l$. Adding the equations, $(a - b) + (b - c) = 3k + 3l$. $a - c = 3(k + l)$. Since $k+l$ is an integer, $a - c$ is divisible by 3. So $aRc$. Transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation on $\mathbb{Z}$.

An equivalence relation partitions the set into disjoint equivalence classes.

The equivalence class of an element $a \in \mathbb{Z}$, denoted by $[a]$, is the set of all integers $x$ such that $xRa$, i.e., $x - a$ is divisible by 3, or $x \equiv a \pmod 3$.

The elements in an equivalence class are integers that have the same remainder when divided by 3.

The possible remainders when an integer is divided by 3 are 0, 1, and 2.

This gives rise to three distinct equivalence classes:

[0] = $\{x \in \mathbb{Z} \mid x \equiv 0 \pmod 3\} = \{..., -6, -3, 0, 3, 6, ...\}$ (the set of integers divisible by 3).

[1] = $\{x \in \mathbb{Z} \mid x \equiv 1 \pmod 3\} = \{..., -5, -2, 1, 4, 7, ...\}$ (the set of integers that leave a remainder of 1 when divided by 3).

[2] = $\{x \in \mathbb{Z} \mid x \equiv 2 \pmod 3\} = \{..., -4, -1, 2, 5, 8, ...\}$ (the set of integers that leave a remainder of 2 when divided by 3).

These three sets [0], [1], and [2] are non-empty, pairwise disjoint, and their union is $\mathbb{Z}$.

These are the pairwise disjoint subsets determined by the equivalence relation R.

The number of such disjoint subsets is the number of possible remainders when dividing by 3, which is 3.

The relation R partitions the set Z into 3 pairwise disjoint subsets.

Given:

Set $\mathbb{R}$: the set of real numbers.

Binary operation $*$ defined on $\mathbb{R}$ as $a * b = a + b - ab$ for all $a, b \in \mathbb{R}$.

To Find:

The identity element with respect to the binary operation $*$.

Solution:

An element $e \in \mathbb{R}$ is the identity element with respect to the binary operation $*$ if for every element $a \in \mathbb{R}$, the following conditions hold:

$a * e = a$ and $e * a = a$.

Using the definition of the operation $*$:

First condition: $a * e = a$

$a + e - ae = a$

Subtract $a$ from both sides:

$e - ae = 0$

Factor out $e$:

$e(1 - a) = 0$

This equation must hold for all $a \in \mathbb{R}$. If this were true for all $a$, it would mean $e=0$. However, the equation $e(1-a)=0$ implies $e=0$ or $1-a=0$ (i.e., $a=1$). This doesn't immediately tell us $e=0$ for all $a$. Let's check the second condition.

Second condition: $e * a = a$

$e + a - ea = a$

Subtract $a$ from both sides:

$e - ea = 0$

Factor out $e$:

$e(1 - a) = 0$

Again, we get $e(1 - a) = 0$. For this equation to hold for all $a \in \mathbb{R}$, the factor multiplying $(1-a)$ must be zero, except possibly for the case when $1-a=0$ (i.e., $a=1$).

Consider $a \neq 1$. Then $1 - a \neq 0$. For $e(1 - a) = 0$ to hold, we must have $e = 0$.

Now let's check if $e=0$ works for all $a \in \mathbb{R}$, including $a=1$.

If $e = 0$, then $a * e = a * 0 = a + 0 - a(0) = a + 0 - 0 = a$.

And $e * a = 0 * a = 0 + a - 0(a) = 0 + a - 0 = a$.

Both conditions $a * 0 = a$ and $0 * a = a$ hold for all $a \in \mathbb{R}$.

Thus, the identity element with respect to the binary operation $*$ is 0.

We can also try to rearrange the expression $a * b = a + b - ab$ to make it easier to find the identity.

Notice that $a + b - ab = 1 - (1 - a)(1 - b)$.

So, $a * b = 1 - (1 - a)(1 - b)$.

Let the identity element be $e$. We need $a * e = a$.

$1 - (1 - a)(1 - e) = a$

$(1 - a)(1 - e) = 1 - a$

$(1 - a)(1 - e) - (1 - a) = 0$

$(1 - a)((1 - e) - 1) = 0$

$(1 - a)(-e) = 0$

$-e(1 - a) = 0$

This implies $e = 0$ or $a = 1$. For this to hold for all $a$, we must have $e=0$.

Check $e * a = a$ using the rewritten form:

$1 - (1 - e)(1 - a) = a$

$1 - (1 - 0)(1 - a) = a$

$1 - 1(1 - a) = a$

$1 - (1 - a) = a$

$1 - 1 + a = a$

$a = a$

This confirms that $e=0$ is the identity element.

The identity element with respect to the binary operation $*$ is 0.

Given:

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

Relation $R = \{(1, 2), (1, 3)\}$ on set A.

Statement: R is a transitive relation.

To Determine:

If the given statement is True or False.

Solution:

A relation $R$ on a set $A$ is transitive if for all $a, b, c \in A$, whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

We need to check all pairs $(a, b) \in R$ and $(b, c) \in R$ to see if the corresponding $(a, c)$ pair is also in $R$.

Let's list the pairs in $R$: $(1, 2)$ and $(1, 3)$.

Case 1: Consider the pair $(a, b) = (1, 2)$. So $a=1, b=2$.

Now we look for pairs in $R$ of the form $(b, c)$, i.e., $(2, c)$.

Are there any pairs in $R$ starting with 2?

Looking at $R = \{(1, 2), (1, 3)\}$, there are no pairs starting with 2. The second elements in $R$ are 2 and 3.

Since there is no pair $(2, c)$ in $R$, the implication "if $(1, 2) \in R$ and $(2, c) \in R$, then $(1, c) \in R$" is vacuously true for $a=1, b=2$. The condition part "$(1, 2) \in R$ and $(2, c) \in R$" is always false because $(2, c) \notin R$ for any $c \in A$. Thus, this part of the transitivity check does not violate the property.

Case 2: Consider the pair $(a, b) = (1, 3)$. So $a=1, b=3$.

Now we look for pairs in $R$ of the form $(b, c)$, i.e., $(3, c)$.

Are there any pairs in $R$ starting with 3?

Looking at $R = \{(1, 2), (1, 3)\}$, there are no pairs starting with 3.

Since there is no pair $(3, c)$ in $R$, the implication "if $(1, 3) \in R$ and $(3, c) \in R$, then $(1, c) \in R$" is vacuously true for $a=1, b=3$. The condition part "$(1, 3) \in R$ and $(3, c) \in R$" is always false because $(3, c) \notin R$ for any $c \in A$. Thus, this part of the transitivity check also does not violate the property.

Since there are no instances where $(a, b) \in R$ and $(b, c) \in R$ for some $a, b, c \in A$, the condition for transitivity (if P then Q) where P is false is always true. Therefore, the relation R satisfies the condition for transitivity.

Alternatively, if the set $S = \{(a, c) \mid \exists b \in A \text{ such that } (a, b) \in R \text{ and } (b, c) \in R\}$ is a subset of $R$, then the relation is transitive.

Let's find the set S for $R = \{(1, 2), (1, 3)\}$.

For $(1, 2) \in R$, we look for pairs $(2, c) \in R$. There are none.

For $(1, 3) \in R$, we look for pairs $(3, c) \in R$. There are none.

Thus, the set $S$ is empty. $S = \phi$.

The condition for transitivity is $S \subseteq R$. Since $\phi$ is a subset of any set, $\phi \subseteq \{(1, 2), (1, 3)\}$.

Therefore, the relation R is transitive.

The statement "R is a transitive relation" is True.

Given:

Set A is a finite set.

Statement: Each injective function from A into itself is not surjective.

To Determine:

If the given statement is True or False.

Solution:

Let A be a finite set with $n$ elements, where $n$ is a non-negative integer. So, $|A| = n$.

Consider a function $f: A \to A$.

We are given that the function $f$ is injective (one-one).

This means that for any distinct elements $x_1, x_2 \in A$, $f(x_1) \neq f(x_2)$. In other words, every element in A is mapped to a unique element in A.

Let's think about the range of this injective function $f$. The range is the set of images of the elements in the domain A.

Range$(f) = \{f(a) \mid a \in A\}$.

Since $f$ is injective, the $n$ distinct elements of A are mapped to $n$ distinct elements in the codomain A.

So, the number of elements in the range is equal to the number of elements in the domain, which is $|A| = n$.

$|$Range$(f)$$| = n$.

The codomain of the function is also A, and $|A| = n$.

Since the number of elements in the range is equal to the number of elements in the codomain ($n$), the range is equal to the codomain.

Range$(f) = A$ (the codomain).

A function is surjective (onto) if its range is equal to its codomain.

Since Range$(f) = A$ (codomain), the function $f$ is surjective.

Therefore, for a finite set A, every injective function from A into itself is also surjective.

The given statement says that each injective function from A into itself is *not* surjective.

This contradicts our finding that for a finite set A, an injective function from A to A is always surjective.

Thus, the statement is false.

Note: This property holds only for functions from a finite set to itself. For infinite sets, an injective function from the set to itself may not be surjective (e.g., $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = 2x$ is injective but not surjective).

The statement "Let A be a finite set. Then, each injective function from A into itself is not surjective" is False.

Given:

Sets A, B, and C.

Functions $f : A \to B$ and $g : B \to C$.

The composite function $g \circ f : A \to C$ is injective.

Statement: Both $f$ and $g$ are injective functions.

To Determine:

If the given statement is True or False.

Solution:

The composite function $g \circ f$ is defined by $(g \circ f)(x) = g(f(x))$ for all $x \in A$.

We are given that $g \circ f$ is injective. This means that for any $x_1, x_2 \in A$, if $(g \circ f)(x_1) = (g \circ f)(x_2)$, then $x_1 = x_2$.

Equivalently, if $g(f(x_1)) = g(f(x_2))$, then $x_1 = x_2$.

Let's analyze the statement that both $f$ and $g$ must be injective.

Is $f$ injective?

Assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in A$.

Apply the function $g$ to both sides of this equality:

$g(f(x_1)) = g(f(x_2))$.

By the definition of the composite function, this is $(g \circ f)(x_1) = (g \circ f)(x_2)$.

Since $g \circ f$ is injective, if $(g \circ f)(x_1) = (g \circ f)(x_2)$, then $x_1 = x_2$.

Thus, starting with $f(x_1) = f(x_2)$, we have shown that $x_1 = x_2$.

This proves that the function $f$ is injective.

Is $g$ injective?

Let's try to prove that $g$ is injective. Suppose $g(y_1) = g(y_2)$ for some $y_1, y_2 \in B$. We want to show that $y_1 = y_2$.

Since $y_1, y_2 \in B$, we can consider if they are in the range of $f$. If $y_1$ and $y_2$ are in the range of $f$, say $y_1 = f(x_1)$ and $y_2 = f(x_2)$ for some $x_1, x_2 \in A$, then $g(f(x_1)) = g(f(x_2))$. Since $g \circ f$ is injective, this implies $x_1 = x_2$, which further implies $f(x_1) = f(x_2)$, so $y_1 = y_2$. This works if $y_1$ and $y_2$ are in the range of $f$.

However, $y_1$ and $y_2$ might be elements in B that are not in the range of $f$. In this case, we cannot necessarily find $x_1, x_2 \in A$ such that $y_1 = f(x_1)$ and $y_2 = f(x_2)$.

Consider a counterexample.

Let $A = \{1\}$, $B = \{a, b\}$, $C = \{c\}$.

Let $f : A \to B$ be defined by $f(1) = a$. This function is injective.

Let $g : B \to C$ be defined by $g(a) = c$ and $g(b) = c$. This function is not injective (since $g(a) = g(b)$ but $a \neq b$).

Now consider the composite function $g \circ f : A \to C$.

$(g \circ f)(1) = g(f(1)) = g(a) = c$.

The function $g \circ f$ maps the single element 1 in A to the single element c in C. Since there is only one element in the domain A, it is impossible for two distinct elements in A to map to the same element in C. Thus, $g \circ f$ is injective.

In this counterexample, $g \circ f$ is injective, and $f$ is injective, but $g$ is not injective.

Therefore, if $g \circ f$ is injective, it is not necessarily true that $g$ is injective.

The statement says that *both* $f$ and $g$ are injective. Since $g$ is not necessarily injective, the statement is false.

The statement "For sets A, B and C, let f : A → B, g : B → C be functions such that g o f is injective. Then both f and g are injective functions" is False.

Given:

Sets A, B, and C.

Functions $f : A \to B$ and $g : B \to C$.

The composite function $g \circ f : A \to C$ is surjective.

Statement: Then $g$ is surjective.

To Determine:

If the given statement is True or False.

Solution:

The composite function $g \circ f$ is defined by $(g \circ f)(x) = g(f(x))$ for all $x \in A$.

We are given that $g \circ f$ is surjective. This means that for every element $z$ in the codomain C, there exists at least one element $x$ in the domain A such that $(g \circ f)(x) = z$.

So, for every $z \in C$, there exists $x \in A$ such that $g(f(x)) = z$.

Let $z$ be an arbitrary element in the codomain C of function $g$.

Since $z \in C$ and $g \circ f : A \to C$ is surjective, there exists an element $x \in A$ such that $(g \circ f)(x) = z$.

By the definition of the composite function, $(g \circ f)(x) = g(f(x))$.

So, $g(f(x)) = z$.

Let $y = f(x)$. Since $x \in A$ and $f : A \to B$, $y = f(x)$ is an element in the codomain of $f$, which is B. So, $y \in B$.

Now we have $g(y) = z$, where $y = f(x)$ and $y \in B$.

We started with an arbitrary element $z$ in the codomain C of function $g$. We found an element $y = f(x)$ in the domain B of function $g$ such that $g(y) = z$.

This means that for every element $z$ in the codomain C of $g$, there exists an element $y$ in the domain B of $g$ such that $g(y) = z$.

This is the definition of a surjective function for $g$.

Therefore, the function $g$ is surjective.

The statement says that if $g \circ f$ is surjective, then $g$ is surjective. Our reasoning above shows that this is true.

Note: It is not necessary for $f$ to be surjective for $g$ to be surjective when $g \circ f$ is surjective. For example, let $A=\{1\}$, $B=\{a, b\}$, $C=\{c\}$. Let $f: A \to B$ be $f(1)=a$. $f$ is not surjective. Let $g: B \to C$ be $g(a)=c, g(b)=c$. $g$ is surjective. $(g \circ f)(1) = g(f(1)) = g(a) = c$. The range of $g \circ f$ is $\{c\}$, and the codomain is $C=\{c\}$. So $g \circ f$ is surjective. In this case, $g$ is surjective as claimed by the statement.

Consider another example where $f$ is not surjective. Let $A = \{1, 2\}$, $B = \{a, b, d\}$, $C = \{c\}$. Let $f: A \to B$ be $f(1)=a, f(2)=b$. $f$ is injective but not surjective (element 'd' in B has no preimage). Let $g: B \to C$ be $g(a)=c, g(b)=c, g(d)=c$. $g$ is surjective. $(g \circ f)(1) = g(f(1)) = g(a) = c$. $(g \circ f)(2) = g(f(2)) = g(b) = c$. The range of $g \circ f$ is $\{c\}$, and the codomain is $C=\{c\}$. So $g \circ f$ is surjective. Here, $g$ is surjective, consistent with the statement.

The statement "For sets A, B and C, let f : A → B, g : B → C be functions such that g o f is surjective. Then g is surjective" is True.

Given:

Set $N = \{1, 2, 3, ...\}$ (the set of natural numbers).

Binary operation $*$ defined on $N$ as $a * b = a + b$ for all $a, b \in N$.

Statement: The binary operation $*$ in $N$ has an identity element.

To Determine:

If the given statement is True or False.

Solution:

An element $e \in N$ is the identity element with respect to the binary operation $*$ if for every element $a \in N$, the following conditions hold:

$a * e = a$ and $e * a = a$.

Using the definition of the operation $*$ ($a * b = a + b$):

First condition: $a * e = a$

$a + e = a$

Subtract $a$ from both sides:

$e = 0$

Second condition: $e * a = a$

$e + a = a$

Subtract $a$ from both sides:

$e = 0$

For the operation $*$ to have an identity element, there must exist an element $e$ such that $e=0$ and $e$ belongs to the set $N$.

The set of natural numbers $N$ is commonly defined as $\{1, 2, 3, ...\}$. In this definition, 0 is not a natural number.

Since the required identity element is 0, and $0 \notin N$, there is no identity element for the operation $+$ on the set of natural numbers $N = \{1, 2, 3, ...\}$.

If the set of natural numbers was defined as $\{0, 1, 2, 3, ...\}$, then 0 would be in the set, and it would be the identity element for the addition operation.

However, given the standard definition of $N$ as $\{1, 2, 3, ...\}$, the element 0 is not in $N$.

Thus, there is no identity element for the binary operation $a * b = a + b$ on the set of natural numbers $N = \{1, 2, 3, ...\}$.

The statement says that the binary operation $*$ in $N$ defined as $a * b = a + b$ has an identity element.

This is false because the required identity element (0) is not in the set $N = \{1, 2, 3, ...\}$.

The statement "Let N be the set of natural numbers. Then, the binary operation * in N defined as a * b = a + b, ∀ a, b ∈ N has identity element" is False.

Given:

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

Relation $R = \{(a, a), (b, c), (a, b)\}$ on set A.

To Find:

The minimum number of ordered pairs to be added to R to make it reflexive and transitive.

Solution:

We need to add pairs to R such that the resulting relation $R'$ is both reflexive and transitive.

Step 1: Ensure Reflexivity

For $R'$ to be reflexive on A, it must contain the pairs $(x, x)$ for all $x \in A$.

The set of reflexive pairs on A is $\{(a, a), (b, b), (c, c)\}$.

The current relation $R$ contains $(a, a)$.

To make it reflexive, we must add the missing reflexive pairs: $(b, b)$ and $(c, c)$.

Let $R_1 = R \cup \{(b, b), (c, c)\} = \{(a, a), (b, c), (a, b), (b, b), (c, c)\}$.

Number of pairs added so far: 2.

Step 2: Ensure Transitivity

Now, consider the relation $R_1$ and check for transitivity. If $(x, y) \in R_1$ and $(y, z) \in R_1$, we must ensure that $(x, z) \in R_1$. We only need to consider pairs where $x \neq z$, as $(x, x)$ is already present for reflexivity.

Let's list the pairs in $R_1$: $(a, a), (b, c), (a, b), (b, b), (c, c)$.

Check pairs of the form $(x, y) \in R_1$ and $(y, z) \in R_1$ with $x \neq z$:

- If $(x, y) = (a, b) \in R_1$, look for pairs $(b, z) \in R_1$ where $z \neq a$.

We have $(b, c) \in R_1$. Here $y=b, z=c$. So we need to check if $(a, c) \in R_1$. Currently, $(a, c) \notin R_1$. We must add $(a, c)$.

We have $(b, b) \in R_1$. Here $y=b, z=b$. $(x, z) = (a, b)$. Since $x \neq z$, we check if $(a, b) \in R_1$. Yes, it is.

- If $(x, y) = (b, c) \in R_1$, look for pairs $(c, z) \in R_1$ where $z \neq b$.

We have $(c, c) \in R_1$. Here $y=c, z=c$. $(x, z) = (b, c)$. Since $x \neq z$, we check if $(b, c) \in R_1$. Yes, it is.

- If $(x, y) = (a, a) \in R_1$, look for pairs $(a, z) \in R_1$ where $z \neq a$.

We have $(a, b) \in R_1$. Here $y=a, z=b$. $(x, z) = (a, b)$. Since $x \neq z$, we check if $(a, b) \in R_1$. Yes, it is.

We have $(a, c) \in R_1$ (which we just identified as needing to be added). If we add $(a, c)$, then with $(a, a) \in R_1$, we need $(a, c) \in R_1$. This is consistent.

- If $(x, y) = (b, b) \in R_1$, look for pairs $(b, z) \in R_1$ where $z \neq b$.

We have $(b, c) \in R_1$. Here $y=b, z=c$. $(x, z) = (b, c)$. Since $x \neq z$, we check if $(b, c) \in R_1$. Yes, it is.

- If $(x, y) = (c, c) \in R_1$, look for pairs $(c, z) \in R_1$ where $z \neq c$.

There are no pairs in $R_1$ starting with $c$ other than $(c, c)$ itself.

Based on the checks, we must add the pair $(a, c)$ to ensure transitivity.

Let's update the relation: $R_2 = R_1 \cup \{(a, c)\} = \{(a, a), (b, c), (a, b), (b, b), (c, c), (a, c)\}$.

Number of pairs added in this step: 1.

Total number of pairs added so far: $2 + 1 = 3$. The added pairs are $(b, b), (c, c), (a, c)$.

Let's verify that $R_2$ is transitive.

List pairs in $R_2$: $(a, a), (b, c), (a, b), (b, b), (c, c), (a, c)$.

Relevant $(x, y), (y, z)$ pairs where $x \neq z$:

- $(a, b) \in R_2$ and $(b, c) \in R_2 \implies (a, c)$ must be in $R_2$. Yes, it is.

- $(a, a) \in R_2$ and $(a, b) \in R_2 \implies (a, b)$ must be in $R_2$. Yes, it is.

- $(a, a) \in R_2$ and $(a, c) \in R_2 \implies (a, c)$ must be in $R_2$. Yes, it is.

- $(b, b) \in R_2$ and $(b, c) \in R_2 \implies (b, c)$ must be in $R_2$. Yes, it is.

- $(c, c) \in R_2$ and no pairs $(c, z)$ with $z \neq c$. Vacuously true.

- $(b, c) \in R_2$ and $(c, c) \in R_2 \implies (b, c)$ must be in $R_2$. Yes, it is.

All necessary transitive pairs are present in $R_2$. $R_2$ is transitive.

By construction, $R_2$ contains all reflexive pairs and was built by adding only the necessary pairs for transitivity based on the current relation at each step.

The pairs added are $(b, b)$, $(c, c)$, and $(a, c)$. The total number of added pairs is 3.

Let's quickly check if $R_2$ is the smallest relation that is reflexive and transitive and contains R. The smallest reflexive relation containing R is $R_1$. The smallest transitive relation containing $R_1$ is found by taking the transitive closure of $R_1$. The transitive closure is the smallest transitive relation containing $R_1$. Our construction essentially finds the transitive closure starting from the reflexive relation $R_1$.

The minimum number of ordered pairs to be added to R to make it reflexive and transitive is 3.

Given:

Real valued function $f(x) = \sqrt{25 - x^2}$.

D is the domain of the function $f$.

To Find:

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

Solution:

For the function $f(x)$ to be a real valued function, the expression under the square root must be non-negative.

So, we must have $25 - x^2 \ge 0$.

We can rearrange this inequality:

$25 \ge x^2$

Equivalently,

$x^2 \le 25$

To solve the inequality $x^2 \le 25$, we consider the roots of $x^2 = 25$, which are $x = \pm 5$.

The inequality $x^2 \le 25$ means that the absolute value of $x$ must be less than or equal to 5.

$|x| \le 5$

This inequality holds for values of $x$ between -5 and 5, inclusive.

So, $-5 \le x \le 5$.

The domain D is the set of all real numbers $x$ that satisfy this condition.

In set-builder notation, the domain is $D = \{x \in \mathbb{R} \mid -5 \le x \le 5\}$.

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

We can also consider the graph of $y = x^2 - 25$. This is a parabola opening upwards, with roots at $x=-5$ and $x=5$. The inequality $25 - x^2 \ge 0$ is equivalent to $-(x^2 - 25) \ge 0$, or $x^2 - 25 \le 0$. This means the parabola $y = x^2 - 25$ must be below or on the x-axis. This occurs between the roots, inclusive.

So, $-5 \le x \le 5$.

The domain D is $[-5, 5]$.

Given:

Function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x + 1$.

Function $g : \mathbb{R} \to \mathbb{R}$ defined by $g(x) = x^2 - 2$.

To Find:

The composite function $g \circ f$.

Solution:

The composite function $g \circ f$ is defined as $g \circ f(x) = g(f(x))$.

We first apply the function $f$ to the input $x$, which gives $f(x) = 2x + 1$.

Then, we apply the function $g$ to the result $f(x)$. The function $g$ takes an input and squares it, then subtracts 2.

So, $g(f(x)) = g(2x + 1)$.

Using the definition of $g(y) = y^2 - 2$ with $y = 2x + 1$, we have:

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

Expand the square:

$(2x + 1)^2 = (2x)^2 + 2(2x)(1) + 1^2 = 4x^2 + 4x + 1$

So, $g(f(x)) = (4x^2 + 4x + 1) - 2$

$g(f(x)) = 4x^2 + 4x - 1$

Therefore, the composite function $g \circ f(x)$ is $4x^2 + 4x - 1$.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x - 3$ for all $x \in \mathbb{R}$.

To Find:

The inverse function $f^{-1}$.

Solution:

To find the inverse function $f^{-1}(x)$, we follow these steps:

1. Let $y = f(x)$.

2. Solve the equation for $x$ in terms of $y$.

3. Replace $y$ with $x$ to get the expression for $f^{-1}(x)$.

Step 1: Let $y = f(x)$.

$y = 2x - 3$

Step 2: Solve for $x$ in terms of $y$.

Add 3 to both sides:

$y + 3 = 2x$

Divide both sides by 2:

$x = \frac{y + 3}{2}$

Step 3: Replace $y$ with $x$ to get $f^{-1}(x)$.

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

The inverse function $f^{-1}$ is defined from $\mathbb{R}$ to $\mathbb{R}$.

The inverse function $f^{-1}(x) = \frac{x + 3}{2}$.

Given:

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

Function $f$ from A to A defined as a set of ordered pairs:

$f = \{(a, b), (b, d), (c, a), (d, c)\}$

To Find:

The inverse function $f^{-1}$.

Solution:

A function $f$ defined as a set of ordered pairs $(x, y)$ means that $f(x) = y$.

For the function $f$ to have an inverse $f^{-1}$, $f$ must be a bijection (one-one and onto).

Let's check if $f$ is a bijection:

- Domain of $f$ is the set of first elements in the ordered pairs: $\{a, b, c, d\}$, which is equal to set A.

- Codomain of $f$ is given implicitly as A.

- Range of $f$ is the set of second elements in the ordered pairs: $\{b, d, a, c\}$, which is equal to set A.

- Injectivity: Each element in the domain is mapped to a unique element in the codomain. The second elements are all distinct: b, d, a, c. So, $f$ is one-one.

- Surjectivity: The range of $f$ is $\{b, d, a, c\}$, which is equal to the codomain A. So, $f$ is onto.

Since $f$ is both one-one and onto, it is a bijection and thus invertible.

The inverse function $f^{-1}$ is obtained by reversing the ordered pairs in the function $f$. If $f(x) = y$, then $f^{-1}(y) = x$. So, if $(x, y) \in f$, then $(y, x) \in f^{-1}$.

The ordered pairs in $f$ are:

$(a, b) \in f \implies f(a) = b$

$(b, d) \in f \implies f(b) = d$

$(c, a) \in f \implies f(c) = a$

$(d, c) \in f \implies f(d) = c$

Reversing these pairs to get the ordered pairs for $f^{-1}$:

If $f(a) = b$, then $f^{-1}(b) = a \implies (b, a) \in f^{-1}$

If $f(b) = d$, then $f^{-1}(d) = b \implies (d, b) \in f^{-1}$

If $f(c) = a$, then $f^{-1}(a) = c \implies (a, c) \in f^{-1}$

If $f(d) = c$, then $f^{-1}(c) = d \implies (c, d) \in f^{-1}$

So, the inverse function $f^{-1}$, as a set of ordered pairs, is:

$f^{-1} = \{(b, a), (d, b), (a, c), (c, d)\}$.

It is standard practice to list the ordered pairs with the first elements in alphabetical order (or the natural order if the domain is ordered). Reordering the pairs:

$f^{-1} = \{(a, c), (b, a), (c, d), (d, b)\}$.

The inverse function $f^{-1} = \{(a, c), (b, a), (c, d), (d, b)\}$.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 - 3x + 2$ for all $x \in \mathbb{R}$.

To Find:

The expression for the composite function $f(f(x))$.

Solution:

The composite function $f(f(x))$ is obtained by substituting $f(x)$ into the function $f$.

Let $y = f(x) = x^2 - 3x + 2$.

Then $f(f(x)) = f(y) = y^2 - 3y + 2$.

Now substitute the expression for $y$ back into the equation:

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

Expand the terms:

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

$= x^4 - 6x^3 + 4x^2 + ( (3x)^2 - 2(3x)(2) + 2^2 )$

$= x^4 - 6x^3 + 4x^2 + 9x^2 - 12x + 4$

$= x^4 - 6x^3 + (4x^2 + 9x^2) - 12x + 4$

$= x^4 - 6x^3 + 13x^2 - 12x + 4$

$-3(x^2 - 3x + 2) = -3x^2 + 9x - 6$

Now combine the expanded terms and the constant +2:

$f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2$

Group like terms:

$f(f(x)) = x^4 - 6x^3 + (13x^2 - 3x^2) + (-12x + 9x) + (4 - 6 + 2)$

$f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x + 0$

$f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x$

The composite function $f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x$.

Given:

A set of ordered pairs $g = \{(1, 1), (2, 3), (3, 5), (4, 7)\}$.

A function described by $g(x) = \alpha x + \beta$.

To Determine:

If $g$ is a function.

The values of $\alpha$ and $\beta$ if $g$ is described by $g(x) = \alpha x + \beta$.

Solution:

Is $g$ a function?

A set of ordered pairs represents a function if and only if no two distinct ordered pairs have the same first element.

The first elements of the ordered pairs in $g$ are 1, 2, 3, and 4.

All the first elements are distinct.

Therefore, $g = \{(1, 1), (2, 3), (3, 5), (4, 7)\}$ is a function.

The domain of this function is $\{1, 2, 3, 4\}$, and the range is $\{1, 3, 5, 7\}$.

Find the values of $\alpha$ and $\beta$:

We are given that the function is described by $g(x) = \alpha x + \beta$.

We can use the given ordered pairs $(x, g(x))$ to form a system of linear equations in $\alpha$ and $\beta$.

Using the pair (1, 1): $g(1) = 1$.

Substitute $x=1$ into $g(x) = \alpha x + \beta$:

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

$1 = \alpha + \beta$ ... (i)

Using the pair (2, 3): $g(2) = 3$.

Substitute $x=2$ into $g(x) = \alpha x + \beta$:

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

$3 = 2\alpha + \beta$ ... (ii)

Now we have a system of two linear equations with two variables $\alpha$ and $\beta$:

Equation (i): $\alpha + \beta = 1$

Equation (ii): $2\alpha + \beta = 3$

Subtract Equation (i) from Equation (ii):

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

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

$\alpha = 2$

Substitute the value of $\alpha = 2$ into Equation (i):

$1 = \alpha + \beta$

$1 = 2 + \beta$

$\beta = 1 - 2$

$\beta = -1$

So, if $g(x) = \alpha x + \beta$, then $\alpha = 2$ and $\beta = -1$.

Let's check if the function $g(x) = 2x - 1$ generates the other ordered pairs:

$g(3) = 2(3) - 1 = 6 - 1 = 5$. This matches the pair (3, 5).

$g(4) = 2(4) - 1 = 8 - 1 = 7$. This matches the pair (4, 7).

Thus, the values $\alpha = 2$ and $\beta = -1$ correctly describe the function $g$.

Yes, g is a function. The values are $\alpha = 2$ and $\beta = -1$.

Given:

Two sets of ordered pairs described by a relation.

To Determine:

For each set:

1. If it represents a function.

2. If it is a function, whether it is injective or surjective.

Solution:

(i) {(x, y): x is a person, y is the mother of x}

This set describes a relation where the domain is the set of all people, and the codomain is also (implicitly) the set of all people. An ordered pair $(x, y)$ is in this set if $x$ is a person and $y$ is the mother of $x$.

Is it a function?

For a set of ordered pairs to be a function, each element in the domain must be associated with exactly one element in the codomain.

In this case, the domain is the set of people. For each person $x$, there is exactly one biological mother $y$. A person cannot have two different mothers simultaneously.

Thus, for every $x$ in the domain (a person), there is a unique $y$ in the codomain (the mother of $x$).

Therefore, this set of ordered pairs is a function.

Is it injective (one-one)?

A function is injective if distinct elements in the domain are mapped to distinct elements in the codomain. This means if $f(x_1) = f(x_2)$, then $x_1 = x_2$. In this context, if two people $x_1$ and $x_2$ have the same mother $y$, are $x_1$ and $x_2$ necessarily the same person?

No. Two different people can have the same mother (e.g., siblings). If $x_1$ and $x_2$ are siblings, they have the same mother $y$, but $x_1 \neq x_2$.

For example, if $x_1$ is Alice and $x_2$ is Bob, and $y$ is Carol (their mother). Then $(Alice, Carol)$ is in the relation, and $(Bob, Carol)$ is in the relation. Here, $Alice \neq Bob$, but $g(Alice) = Carol$ and $g(Bob) = Carol$.

Therefore, the function is not injective.

Is it surjective (onto)?

A function is surjective if every element in the codomain has at least one preimage in the domain. The codomain is the set of all people.

Does every person $y$ in the codomain have a person $x$ in the domain such that $y$ is the mother of $x$? In other words, is every person a mother?

No. Men are not mothers. Also, some women may not be mothers.

Therefore, there are elements in the codomain (people who are not mothers) that do not have a preimage in the domain.

The function is not surjective.

(ii) {(a, b): a is a person, b is an ancestor of a}

This set describes a relation where the domain is the set of all people, and the codomain is also (implicitly) the set of all people. An ordered pair $(a, b)$ is in this set if $a$ is a person and $b$ is an ancestor of $a$. An ancestor typically includes parents, grandparents, great-grandparents, etc., but not the person itself.

Is it a function?

For a set of ordered pairs to be a function, each element in the domain must be associated with exactly one element in the codomain.

In this case, the domain is the set of people. For a person $a$, is there exactly one ancestor $b$?

No. Every person has multiple ancestors (two parents, four grandparents, eight great-grandparents, and so on, excluding duplicates from intermarriage). For a given person $a$, there are many people $b$ who are ancestors of $a$.

For example, if $a$ is a person, $a$'s father is an ancestor, and $a$'s mother is an ancestor. If the relation contains $(a, father)$ and $(a, mother)$, and father $\neq$ mother, then the element $a$ in the domain is associated with two different elements in the codomain. This violates the condition for being a function.

Therefore, this set of ordered pairs is not a function.

Since it is not a function, we do not need to examine injectivity or surjectivity.

Summary:

(i) {(x, y): x is a person, y is the mother of x} is a function. It is neither injective nor surjective.

(ii) {(a, b): a is a person, b is an ancestor of a} is not a function.

Given:

Function $f$ as a set of ordered pairs: $f = \{(1, 2), (3, 5), (4, 1)\}$.

Function $g$ as a set of ordered pairs: $g = \{(2, 3), (5, 1), (1, 3)\}$.

To Find:

The composite function $f \circ g$.

Solution:

The composite function $f \circ g$ is defined as $f \circ g(x) = f(g(x))$.

The domain of $g$ is the set of first elements in the ordered pairs of $g$: Domain$(g) = \{2, 5, 1\}$.

The domain of $f$ is the set of first elements in the ordered pairs of $f$: Domain$(f) = \{1, 3, 4\}$.

The range of $g$ is the set of second elements in the ordered pairs of $g$: Range$(g) = \{3, 1\}$.

The range of $f$ is the set of second elements in the ordered pairs of $f$: Range$(f) = \{2, 5, 1\}$.

The composite function $f \circ g$ is defined for values of $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.

Domain$(f \circ g) = \{x \in \text{Domain}(g) \mid g(x) \in \text{Domain}(f)\}$.

Let's evaluate $f(g(x))$ for each element $x$ in the domain of $g$ and check if $g(x)$ is in the domain of $f$.

For $x = 2 \in \text{Domain}(g)$:

$g(2) = 3$. Is $g(2) = 3$ in Domain$(f) = \{1, 3, 4\}$?

Yes, 3 is in Domain$(f)$. So, we can evaluate $f(g(2))$.

$f(g(2)) = f(3)$. From $f = \{(1, 2), (3, 5), (4, 1)\}$, we have $f(3) = 5$.

So, $(2, 5)$ is an ordered pair in $f \circ g$.

For $x = 5 \in \text{Domain}(g)$:

$g(5) = 1$. Is $g(5) = 1$ in Domain$(f) = \{1, 3, 4\}$?

Yes, 1 is in Domain$(f)$. So, we can evaluate $f(g(5))$.

$f(g(5)) = f(1)$. From $f = \{(1, 2), (3, 5), (4, 1)\}$, we have $f(1) = 2$.

So, $(5, 2)$ is an ordered pair in $f \circ g$.

For $x = 1 \in \text{Domain}(g)$:

$g(1) = 3$. Is $g(1) = 3$ in Domain$(f) = \{1, 3, 4\}$?

Yes, 3 is in Domain$(f)$. So, we can evaluate $f(g(1))$.

$f(g(1)) = f(3)$. From $f = \{(1, 2), (3, 5), (4, 1)\}$, we have $f(3) = 5$.

So, $(1, 5)$ is an ordered pair in $f \circ g$.

The ordered pairs in the composite function $f \circ g$ are the pairs $(x, f(g(x)))$ that we found.

$f \circ g = \{(2, 5), (5, 2), (1, 5)\}$.

It is conventional to list the pairs by the order of the first element. Reordering the pairs:

$f \circ g = \{(1, 5), (2, 5), (5, 2)\}$.

The domain of $f \circ g$ is $\{1, 2, 5\}$.

The composite function $f \circ g = \{(1, 5), (2, 5), (5, 2)\}$.

Given:

Set $\mathbb{C}$: the set of complex numbers.

Set $\mathbb{R}$: the set of real numbers.

Function $f : \mathbb{C} \to \mathbb{R}$ defined by $f(z) = |z|$ for all $z \in \mathbb{C}$. The modulus of a complex number $z = x + iy$ (where $x, y \in \mathbb{R}$) is defined as $|z| = \sqrt{x^2 + y^2}$.

To Prove:

The function $f$ is neither one-one (injective) nor onto (surjective).

Proof:

Not One-one (Injectivity):

A function $f$ is one-one if for any $z_1, z_2$ in the domain $\mathbb{C}$, $f(z_1) = f(z_2)$ implies $z_1 = z_2$.

To show that $f$ is not one-one, we need to find two distinct complex numbers $z_1, z_2 \in \mathbb{C}$ such that $f(z_1) = f(z_2)$.

Let $z_1 = 1$. $z_1$ is a complex number (it can be written as $1 + 0i$).

$f(z_1) = |1| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1$.

Let $z_2 = -1$. $z_2$ is a complex number (it can be written as $-1 + 0i$). Note that $z_1 \neq z_2$.

$f(z_2) = |-1| = \sqrt{(-1)^2 + 0^2} = \sqrt{1} = 1$.

We have $f(z_1) = f(z_2) = 1$, but $z_1 = 1$ and $z_2 = -1$ are distinct complex numbers ($1 \neq -1$).

As another example, consider $z_1 = i$ and $z_2 = -i$. $i$ is a complex number ($0 + 1i$), $-i$ is a complex number ($0 - 1i$). $i \neq -i$.

$f(i) = |i| = \sqrt{0^2 + 1^2} = \sqrt{1} = 1$.

$f(-i) = |-i| = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1$.

We have $f(i) = f(-i) = 1$, but $i \neq -i$.

In general, any two distinct complex numbers that lie on the same circle centered at the origin in the complex plane have the same modulus. For example, $z_1 = 1 + i$ and $z_2 = \sqrt{2}$.

$f(1+i) = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

$f(\sqrt{2}) = |\sqrt{2}| = \sqrt{(\sqrt{2})^2 + 0^2} = \sqrt{2}$.

However, $1+i \neq \sqrt{2}$.

Since we can find distinct complex numbers that have the same image under $f$, the function $f$ is not one-one.

Not Onto (Surjectivity):

A function $f : A \to B$ is onto if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$.

The codomain of $f$ is given as $\mathbb{R}$. To show that $f$ is not onto, we need to find a real number $y \in \mathbb{R}$ such that there is no complex number $z \in \mathbb{C}$ for which $f(z) = y$.

The definition of $f(z)$ is $|z| = \sqrt{x^2 + y^2}$, where $z = x + iy$ and $x, y \in \mathbb{R}$.

The term $x^2$ is always non-negative ($x^2 \ge 0$) because $x$ is a real number.

The term $y^2$ is always non-negative ($y^2 \ge 0$) because $y$ is a real number.

The sum of two non-negative numbers is non-negative: $x^2 + y^2 \ge 0$.

The square root of a non-negative number is always non-negative. So, $\sqrt{x^2 + y^2} \ge 0$.

This means that the value of $f(z) = |z|$ is always a non-negative real number.

$f(z) \ge 0$ for all $z \in \mathbb{C}$.

The range of the function $f$ is the set of all non-negative real numbers, $[0, \infty)$.

The codomain of $f$ is given as $\mathbb{R}$. However, the range of $f$ is $[0, \infty)$, which is a proper subset of the codomain $\mathbb{R}$.

For example, consider a negative real number from the codomain, such as $y = -5$. The value $-5$ is in the codomain $\mathbb{R}$. However, $-5$ is not in the range $[0, \infty)$. This means there is no complex number $z$ such that $f(z) = |z| = -5$.

Since there are elements in the codomain $\mathbb{R}$ (like $y=-5$) for which there is no corresponding element in the domain $\mathbb{C}$, the function $f$ is not onto.

Therefore, the function $f(z) = |z|$ defined from the set of complex numbers $\mathbb{C}$ to the set of real numbers $\mathbb{R}$ is neither one-one nor onto.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \cos x$ for all $x \in \mathbb{R}$.

To Show:

The function $f$ is neither one-one (injective) nor onto (surjective).

Proof:

Not One-one (Injectivity):

A function $f$ is one-one if for any $x_1, x_2$ in the domain $\mathbb{R}$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

To show that $f$ is not one-one, we need to find two distinct real numbers $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$.

Consider the values of $\cos x$ for different $x$. The cosine function is periodic with a period of $2\pi$. This means $\cos(x) = \cos(x + 2\pi)$ for any real number $x$.

Let $x_1 = 0$. Then $f(x_1) = f(0) = \cos(0) = 1$.

Let $x_2 = 2\pi$. Then $f(x_2) = f(2\pi) = \cos(2\pi) = 1$.

We have $f(0) = f(2\pi) = 1$, but $0 \neq 2\pi$. Thus, the function is not one-one.

In general, for any $x \in \mathbb{R}$, $\cos(x) = \cos(x + 2n\pi)$ for any integer $n$. We can always find infinitely many distinct values of $x$ that have the same cosine value.

Therefore, the function $f(x) = \cos x$ is not one-one.

Not Onto (Surjectivity):

A function $f : A \to B$ is onto if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$.

The codomain of $f$ is given as $\mathbb{R}$. To show that $f$ is not onto, we need to find a real number $y \in \mathbb{R}$ such that there is no real number $x \in \mathbb{R}$ for which $f(x) = y$.

The range of the cosine function, $\cos x$, is well-known to be the interval $[-1, 1]$. This means that for any real number $x$, the value of $\cos x$ is always between -1 and 1, inclusive.

$-1 \le \cos x \le 1$ for all $x \in \mathbb{R}$.

The range of the function $f(x) = \cos x$ is $[-1, 1]$.

The codomain of $f$ is given as $\mathbb{R}$. However, the range of $f$ is $[-1, 1]$, which is a proper subset of the codomain $\mathbb{R}$.

For example, consider a real number from the codomain that is not in the range, such as $y = 2$. The value $2$ is in the codomain $\mathbb{R}$. However, $2$ is not in the interval $[-1, 1]$. This means there is no real number $x$ such that $f(x) = \cos x = 2$.

As another example, consider $y = -3$. This is in the codomain $\mathbb{R}$, but not in the range $[-1, 1]$. There is no real number $x$ such that $\cos x = -3$.

Since there are elements in the codomain $\mathbb{R}$ (like $y=2$ or $y=-3$) for which there is no corresponding element in the domain $\mathbb{R}$, the function $f$ is not onto.

Therefore, the function $f(x) = \cos x$ defined from $\mathbb{R}$ to $\mathbb{R}$ is neither one-one nor onto.

Given:

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

Set $Y = \{4, 5\}$.

Four subsets of $X \times Y$.

To Determine:

Which of the given subsets of $X \times Y$ are functions from $X$ to $Y$.

Solution:

A subset $R$ of $X \times Y$ is a function from $X$ to $Y$ if and only if two conditions are met:

1. Every element in the domain $X$ is the first element of at least one ordered pair in the subset $R$.

2. No two distinct ordered pairs in the subset $R$ have the same first element.

Let's check each subset:

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

Domain check: The first elements are 1, 1, 2, 3. The elements of $X$ are 1, 2, 3. All elements in $X$ appear as the first element of at least one pair.

Uniqueness check: Consider the first element 1. There are two ordered pairs with 1 as the first element: $(1, 4)$ and $(1, 5)$. Since these are two distinct ordered pairs with the same first element (and different second elements), this violates the condition for being a function.

Therefore, $f$ is not a function from $X$ to $Y$.

(ii) $g = \{(1, 4), (2, 4), (3, 4)\}

Domain check: The first elements are 1, 2, 3. The elements of $X$ are 1, 2, 3. All elements in $X$ appear as the first element of exactly one pair. Condition 1 is met (since each appears exactly once, each appears at least once).

Uniqueness check: The first elements are 1, 2, 3. All the first elements are distinct. Condition 2 is met.

Therefore, $g$ is a function from $X$ to $Y$.

Note: In this function, different elements in the domain are mapped to the same element in the codomain. This is allowed for a function, but means it is not injective.

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

Domain check: The first elements are 1, 2, 3. The elements of $X$ are 1, 2, 3. All elements in $X$ appear as the first element of exactly one pair. Condition 1 is met.

Uniqueness check: The first elements are 1, 2, 3. All the first elements are distinct. Condition 2 is met.

Therefore, $h$ is a function from $X$ to $Y$.

Note: In this function, elements 2 and 3 are mapped to the same element 5. This is allowed for a function, but means it is not injective.

(iv) $k = \{(1, 4), (2, 5)\}

Domain check: The first elements are 1, 2. The elements of $X$ are 1, 2, 3. The element 3 in $X$ does not appear as the first element of any ordered pair in $k$. This violates the condition that every element in the domain must be the first element of at least one ordered pair.

Therefore, $k$ is not a function from $X$ to $Y$. (It is a function from $\{1, 2\}$ to $Y$).

Summary:

(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)}: Not a function.

(ii) g = {(1, 4), (2, 4), (3, 4)}: Is a function.

(iii) h = {(1,4), (2, 5), (3, 5)}: Is a function.

(iv) k = {(1,4), (2, 5)}: Not a function.

Given:

Functions $f : A \to B$ and $g : B \to A$.

The composite function $g \circ f = I_A$, where $I_A$ is the identity function on set A. This means $(g \circ f)(x) = x$ for all $x \in A$.

To Show:

The function $f$ is one-one (injective).

The function $g$ is onto (surjective).

Proof:

Show that f is one-one:

A function $f : A \to B$ is one-one if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in A$.

Apply the function $g$ to both sides of this equality:

$g(f(x_1)) = g(f(x_2))$

By the definition of the composite function, $g(f(x)) = (g \circ f)(x)$. So we have:

$(g \circ f)(x_1) = (g \circ f)(x_2)$

We are given that $g \circ f = I_A$. The identity function $I_A$ maps each element in A to itself. So, $(g \circ f)(x) = x$ for all $x \in A$.

Therefore, $(g \circ f)(x_1) = x_1$ and $(g \circ f)(x_2) = x_2$.

Substituting these into the equation above:

$x_1 = x_2$

Thus, starting with $f(x_1) = f(x_2)$, we have shown that $x_1 = x_2$.

This proves that the function $f$ is one-one (injective).

Show that g is onto:

A function $g : B \to A$ is onto if for every element $x$ in the codomain A, there exists at least one element $y$ in the domain B such that $g(y) = x$.

Let $x$ be an arbitrary element in the codomain A of function $g$.

We need to find an element $y \in B$ such that $g(y) = x$.

We are given the condition $(g \circ f)(x) = x$ for all $x \in A$.

Let's consider an element $x \in A$. Applying the definition of $g \circ f$, we have $g(f(x)) = x$.

Let $y = f(x)$. Since $x \in A$ and $f : A \to B$, $y = f(x)$ is an element in the codomain of $f$, which is B. So, $y \in B$.

Now we have $g(y) = x$, where $y = f(x)$ and $y \in B$.

We started with an arbitrary element $x$ in the codomain A of function $g$. We found an element $y = f(x)$ in the domain B of function $g$ such that $g(y) = x$.

This means that for every element $x$ in the codomain A of $g$, there exists an element $y$ in the domain B of $g$ such that $g(y) = x$.

This is the definition of a surjective function for $g$.

Therefore, the function $g$ is onto (surjective).

Thus, if $f : A \to B$ and $g : B \to A$ satisfy $g \circ f = I_A$, then $f$ is one-one and $g$ is onto.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{1}{2 - \cos x}$ for all $x \in \mathbb{R}$.

To Find:

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

Solution:

The domain of the function is $\mathbb{R}$. We know the range of the cosine function is $[-1, 1]$.

So, for any real number $x$, we have:

$-1 \le \cos x \le 1$

We want to find the range of $f(x) = \frac{1}{2 - \cos x}$. Let's analyze the denominator, $2 - \cos x$.

Multiply the inequality by -1 and reverse the inequality signs:

$1 \ge -\cos x \ge -1$

or $-1 \le -\cos x \le 1$.

Now, add 2 to all parts of the inequality:

$2 + (-1) \le 2 - \cos x \le 2 + 1$

$1 \le 2 - \cos x \le 3$

So, the denominator $2 - \cos x$ takes values in the interval $[1, 3]$.

Since $1 \le 2 - \cos x \le 3$, the denominator is never zero, so the function is well-defined for all $x \in \mathbb{R}$.

Now consider the reciprocal $\frac{1}{2 - \cos x}$. When taking the reciprocal of a positive interval $[a, b]$ where $0 < a \le b$, the new interval is $[\frac{1}{b}, \frac{1}{a}]$.

Here, $a=1$ and $b=3$. So, the reciprocal values $\frac{1}{2 - \cos x}$ will lie in the interval $\left[\frac{1}{3}, \frac{1}{1}\right]$.

$\frac{1}{3} \le \frac{1}{2 - \cos x} \le 1$

This means that the values of $f(x)$ are in the interval $\left[\frac{1}{3}, 1\right]$.

We need to confirm that all values in this interval are attained. Since $\cos x$ can take any value in $[-1, 1]$, the denominator $2 - \cos x$ can take any value in $[1, 3]$. As the reciprocal function $y = \frac{1}{u}$ is continuous and decreasing for $u > 0$, it maps the interval $[1, 3]$ to the interval $\left[\frac{1}{3}, 1\right]$.

So, the range of the function $f(x)$ is the interval $\left[\frac{1}{3}, 1\right]$.

The range of the function $f(x)$ is $\left[\frac{1}{3}, 1\right]$.

Given:

Set $\mathbb{Z}$: the set of integers.

$n$: a fixed positive integer.

Relation $R$ on $\mathbb{Z}$ defined by $aRb$ if and only if $a - b$ is divisible by $n$, for all $a, b \in \mathbb{Z}$.

To Show:

The relation $R$ is an equivalence relation.

Proof:

For a relation R to be an equivalence relation on a set, it must satisfy the following three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

The relation is given by $aRb \iff a - b$ is divisible by $n$. This can be written using modular arithmetic notation as $a \equiv b \pmod n$.

1. Reflexivity:

A relation $R$ on a set $\mathbb{Z}$ is reflexive if $aRa$ for every element $a \in \mathbb{Z}$.

We need to check if $aRa$ holds for any $a \in \mathbb{Z}$. According to the definition of R, $aRa$ if and only if $a - a$ is divisible by $n$.

Consider $a - a = 0$.

Is 0 divisible by $n$ for a positive integer $n$?

Yes, $0 = n \times 0$. Since 0 is an integer, 0 is divisible by $n$.

So, $a - a$ is divisible by $n$ for all $a \in \mathbb{Z}$.

Thus, $aRa$ for all $a \in \mathbb{Z}$. The relation $R$ is reflexive.

2. Symmetry:

A relation $R$ on a set $\mathbb{Z}$ is symmetric if whenever $aRb$, then $bRa$ for all $a, b \in \mathbb{Z}$.

Assume $aRb$ for some $a, b \in \mathbb{Z}$.

According to the definition of R, $aRb$ means $a - b$ is divisible by $n$.

This implies that there exists an integer $k$ such that $a - b = nk$.

Now consider $b - a$. We can write $b - a = -(a - b)$.

Substituting $a - b = nk$, we get $b - a = -(nk) = n(-k)$.

Since $k$ is an integer, $-k$ is also an integer.

So, $b - a$ is an integer multiple of $n$. This means $b - a$ is divisible by $n$.

According to the definition of R, this means $bRa$.

Thus, if $aRb$, then $bRa$. The relation $R$ is symmetric.

3. Transitivity:

A relation $R$ on a set $\mathbb{Z}$ is transitive if whenever $aRb$ and $bRc$, then $aRc$ for all $a, b, c \in \mathbb{Z}$.

Assume $aRb$ and $bRc$ for some $a, b, c \in \mathbb{Z}$.

According to the definition of R:

$aRb$ means $a - b$ is divisible by $n$. This implies $a - b = nk$ for some integer $k$.

$bRc$ means $b - c$ is divisible by $n$. This implies $b - c = nl$ for some integer $l$.

We want to check if $aRc$, which means $a - c$ is divisible by $n$.

Consider the sum of the differences $(a - b) + (b - c)$.

$(a - b) + (b - c) = a - b + b - c = a - c$.

Substituting the expressions from the given conditions:

$a - c = (nk) + (nl)$

$a - c = n(k + l)$

Since $k$ and $l$ are integers, their sum $k + l$ is also an integer.

So, $a - c$ is an integer multiple of $n$. This means $a - c$ is divisible by $n$.

According to the definition of R, this means $aRc$.

Thus, if $aRb$ and $bRc$, then $aRc$. The relation $R$ is transitive.

Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation on the set of integers $\mathbb{Z}$.

The relation R defined by $aRb$ if and only if $a - b$ is divisible by $n$ is an equivalence relation on $\mathbb{Z}$.

Given set $A = \{1, 2, 3, 4\}$. We need to define relations on $A$ based on the given properties.

(a) Reflexive, transitive but not symmetric

A relation $R$ on $A$ is:

• Reflexive if $(a, a) \in R$ for all $a \in A$.
• Symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$.
• Transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$ for all $a, b, c \in A$.

We need a relation $R_1$ such that it satisfies reflexivity and transitivity, but not symmetry.

To ensure reflexivity, $R_1$ must contain the pairs $(1,1), (2,2), (3,3), (4,4)$.

To ensure it is not symmetric, we must include a pair $(a,b)$ where $a \neq b$, but not include the pair $(b,a)$. Let's include $(1,2)$.

So far, $R_1$ contains $\{(1,1), (2,2), (3,3), (4,4), (1,2)\}$.

Now, we need to check for transitivity and add pairs if necessary to satisfy it. If $(a,b) \in R_1$ and $(b,c) \in R_1$, then $(a,c)$ must be in $R_1$. Consider the pairs in our current $R_1$:

$(1,2) \in R_1$. Are there any pairs starting with 2? Only $(2,2)$. $(1,2) \in R_1$ and $(2,2) \in R_1 \implies (1,2) \in R_1$ (already included).

Consider other pairs: $(1,1), (2,2), (3,3), (4,4)$. $(1,1) \in R_1$ and $(1,2) \in R_1 \implies (1,2) \in R_1$ (already included).

All pairs of the form $(a,a)$ and $(a,a)$ are transitive as $(a,a) \in R_1$. All pairs of the form $(a,a)$ and $(a,b)$ are transitive as $(a,b) \in R_1$. All pairs of the form $(b,c)$ and $(c,c)$ are transitive as $(b,c) \in R_1$.

The only non-reflexive pair is $(1,2)$. We checked cases involving $(1,2)$. No other pairs are needed to satisfy transitivity with the current elements.

Let's consider a slightly larger relation to make the non-symmetry more obvious or provide another example. Consider $R_1 = \{(1,1), (2,2), (3,3), (4,4), (1,2), (2,3), (1,3)\}$.

Reflexive: Yes, $(1,1), (2,2), (3,3), (4,4)$ are present.

Symmetric: No, $(1,2) \in R_1$ but $(2,1) \notin R_1$. Also $(2,3) \in R_1$ but $(3,2) \notin R_1$, and $(1,3) \in R_1$ but $(3,1) \notin R_1$.

Transitive: Check all combinations $(a,b), (b,c) \in R_1$:

• $(1,1)$ with any pair starting with 1: $(1,1), (1,2) \implies (1,2) \in R_1$. $(1,1), (1,3) \implies (1,3) \in R_1$.
• $(2,2)$ with any pair starting with 2: $(2,2), (2,3) \implies (2,3) \in R_1$.
• $(3,3)$ with any pair starting with 3: None except $(3,3)$ itself.
• $(4,4)$ with any pair starting with 4: None except $(4,4)$ itself.
• $(1,2)$ with any pair starting with 2: $(1,2), (2,2) \implies (1,2) \in R_1$. $(1,2), (2,3) \implies (1,3) \in R_1$.
• $(2,3)$ with any pair starting with 3: $(2,3), (3,3) \implies (2,3) \in R_1$.
• $(1,3)$ with any pair starting with 3: $(1,3), (3,3) \implies (1,3) \in R_1$.

Thus, an example of a relation on $A$ that is reflexive, transitive but not symmetric is:

$R_1 = \{(1,1), (2,2), (3,3), (4,4), (1,2), (2,3), (1,3)\}$.

(b) Symmetric but neither reflexive nor transitive

We need a relation $R_2$ such that it is symmetric, but not reflexive and not transitive.

To ensure it is not reflexive, we must exclude at least one pair $(a,a)$ from $R_2$. The simplest way is to exclude all of them.

To ensure it is symmetric, if $(a,b) \in R_2$ for $a \neq b$, then $(b,a)$ must also be in $R_2$. Let's include $(1,2)$ and its symmetric pair $(2,1)$.

So far, $R_2$ contains $\{(1,2), (2,1)\}$.

Now, let's check the properties:

Reflexive: No, $(1,1), (2,2), (3,3), (4,4)$ are not in $R_2$.

Symmetric: Yes, $(1,2) \in R_2 \implies (2,1) \in R_2$. $(2,1) \in R_2 \implies (1,2) \in R_2$. There are no other pairs $(a,b)$ with $a \neq b$, so the condition holds.

Transitive: Check if for every $(a,b) \in R_2$ and $(b,c) \in R_2$, $(a,c) \in R_2$. Consider $(1,2) \in R_2$. A pair starting with 2 is $(2,1)$. $(1,2) \in R_2$ and $(2,1) \in R_2 \implies (1,1)$ must be in $R_2$. But $(1,1) \notin R_2$. Thus, the relation is not transitive.

Thus, an example of a relation on $A$ that is symmetric but neither reflexive nor transitive is:

$R_2 = \{(1,2), (2,1)\}$.

(c) Reflexive, symmetric and transitive

A relation that is reflexive, symmetric, and transitive is called an equivalence relation.

We need a relation $R_3$ such that it is reflexive, symmetric, and transitive.

To ensure reflexivity, $R_3$ must contain the pairs $(1,1), (2,2), (3,3), (4,4)$.

Let's consider the relation containing only these pairs:

$R_3 = \{(1,1), (2,2), (3,3), (4,4)\}$.

Let's check the properties for $R_3$:

Reflexive: Yes, $(a,a) \in R_3$ for all $a \in \{1,2,3,4\}$.

Symmetric: Yes. If $(a,b) \in R_3$, then it must be of the form $(a,a)$. The symmetric pair is $(a,a)$ itself, which is in $R_3$.

Transitive: Yes. If $(a,b) \in R_3$ and $(b,c) \in R_3$, then $(a,b)$ must be of the form $(a,a)$ (i.e., $a=b$) and $(b,c)$ must be of the form $(b,b)$ (i.e., $b=c$). This implies $a=b=c$. So $(a,c) = (a,a)$, which is in $R_3$.

Thus, an example of a relation on $A$ that is reflexive, symmetric and transitive is:

$R_3 = \{(1,1), (2,2), (3,3), (4,4)\}$.

This relation corresponds to the partition of $A$ into singleton sets: $\{1\}, \{2\}, \{3\}, \{4\}$.

Given:

Set $N$ of natural numbers, $N = \{1, 2, 3, \ldots\}$.

Relation $R$ defined on $N$ as $R = \{(x, y): x \in N, y \in N, 2x + y = 41\}$.

To Find:

The domain and range of $R$.

Whether $R$ is reflexive, symmetric, and transitive.

Solution:

The relation $R$ consists of ordered pairs $(x, y)$ of natural numbers satisfying the equation $2x + y = 41$. We can rewrite the equation as $y = 41 - 2x$.

Let's find the pairs $(x, y)$ in $R$ by substituting values for $x \in N$ and checking if the resulting $y$ is also in $N$.

• If $x=1$, $y = 41 - 2(1) = 39$. $(1, 39) \in R$ since $1 \in N, 39 \in N$.
• If $x=2$, $y = 41 - 2(2) = 37$. $(2, 37) \in R$ since $2 \in N, 37 \in N$.
• If $x=3$, $y = 41 - 2(3) = 35$. $(3, 35) \in R$ since $3 \in N, 35 \in N$.
• ...
• We continue this until $y$ is no longer a natural number. Since $y \in N$, $y \ge 1$.
• $41 - 2x \ge 1$
• $40 \ge 2x$
• $20 \ge x$
• Since $x \in N$, the possible values for $x$ are $1, 2, 3, \ldots, 20$.
• If $x=20$, $y = 41 - 2(20) = 41 - 40 = 1$. $(20, 1) \in R$ since $20 \in N, 1 \in N$.
• If $x=21$, $y = 41 - 2(21) = 41 - 42 = -1$. $(-1)$ is not a natural number, so no pairs with $x \ge 21$ are in $R$.

The set of pairs in $R$ is $\{(1, 39), (2, 37), (3, 35), \ldots, (20, 1)\}$.

Domain of R:

The domain of $R$ is the set of all first elements $x$ in the ordered pairs $(x, y) \in R$.

From the pairs we found, the possible values for $x$ are $1, 2, 3, \ldots, 20$.

Domain$(R) = \{x \in N : \exists y \in N \text{ such that } 2x + y = 41\}$.

We found that this set is $\{1, 2, 3, \ldots, 20\}$.

Domain$(R) = \{1, 2, 3, \ldots, 20\}$.

Range of R:

The range of $R$ is the set of all second elements $y$ in the ordered pairs $(x, y) \in R$.

The possible values for $y$ are the corresponding values when $x = 1, 2, \ldots, 20$.

• If $x=1$, $y=39$.
• If $x=2$, $y=37$.
• If $x=3$, $y=35$.
• ...
• If $x=20$, $y=1$.

The values of $y$ are $39, 37, 35, \ldots, 1$. These are the odd natural numbers from 1 to 39.

Range$(R) = \{y \in N : \exists x \in N \text{ such that } 2x + y = 41\}$.

We found that this set is $\{1, 3, 5, \ldots, 39\}$.

Range$(R) = \{1, 3, 5, \ldots, 39\}$.

Verification of Properties:

Reflexivity:

A relation $R$ on a set $A$ is reflexive if $(a, a) \in R$ for every $a \in A$. Here $A = N$.

For $(a, a)$ to be in $R$, it must satisfy $2a + a = 41$, which means $3a = 41$.

Solving for $a$, we get $a = \frac{41}{3}$.

Since $\frac{41}{3}$ is not a natural number, there is no $a \in N$ such that $(a, a) \in R$.

For example, consider $a=1 \in N$. $(1, 1) \notin R$ because $2(1) + 1 = 3 \neq 41$.

Therefore, the relation $R$ is not reflexive.

Symmetry:

A relation $R$ on a set $A$ is symmetric if for all $a, b \in A$, if $(a, b) \in R$, then $(b, a) \in R$. Here $A = N$.

Suppose $(x, y) \in R$. This means $x \in N, y \in N$, and $2x + y = 41$.

For $R$ to be symmetric, if $(x, y) \in R$, then $(y, x)$ must also be in $R$. For $(y, x)$ to be in $R$, it must satisfy $2y + x = 41$.

Let's check with a pair from $R$. We know that $(1, 39) \in R$ because $1 \in N, 39 \in N$ and $2(1) + 39 = 41$.

Now let's check if $(39, 1) \in R$. For $(39, 1)$ to be in $R$, it must satisfy $2(39) + 1 = 41$.

$2(39) + 1 = 78 + 1 = 79$.

Since $79 \neq 41$, the pair $(39, 1) \notin R$.

We found a pair $(1, 39) \in R$ such that $(39, 1) \notin R$.

Therefore, the relation $R$ is not symmetric.

Transitivity:

A relation $R$ on a set $A$ is transitive if for all $a, b, c \in A$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. Here $A = N$.

Suppose $(x, y) \in R$ and $(y, z) \in R$. This means:

• $x \in N, y \in N$, and $2x + y = 41$.
• $y \in N, z \in N$, and $2y + z = 41$.

We need to check if $(x, z) \in R$, i.e., if $2x + z = 41$ must hold.

From $2x + y = 41$, we have $y = 41 - 2x$.

From $2y + z = 41$, we have $z = 41 - 2y$.

Substitute the first equation into the second for $y$:

$z = 41 - 2(41 - 2x)$

$z = 41 - 82 + 4x$

$z = 4x - 41$

For $(x, z)$ to be in $R$, we require $2x + z = 41$. Substitute the expression for $z$:

$2x + (4x - 41) = 41$

$6x - 41 = 41$

$6x = 82$

$x = \frac{82}{6} = \frac{41}{3}$.

This implies that transitivity would only hold if $x$ must be $\frac{41}{3}$, which is not possible since $x$ must be a natural number. This line of reasoning shows that if $x \neq 41/3$, transitivity fails unless no such chains $(x,y), (y,z)$ exist.

Let's find a specific counterexample.

We need $(x, y) \in R$ and $(y, z) \in R$ for some $x, y, z \in N$. From our list of pairs, $(11, 19) \in R$ because $11 \in N, 19 \in N$ and $2(11) + 19 = 22 + 19 = 41$.

Now, consider a pair starting with $y=19$. We need $(19, z) \in R$ for some $z \in N$. For $(19, z) \in R$, we must have $19 \in N, z \in N$ and $2(19) + z = 41$. $38 + z = 41$ $z = 41 - 38 = 3$. Since $3 \in N$, the pair $(19, 3)$ is in $R$. So we have $(11, 19) \in R$ and $(19, 3) \in R$.

For $R$ to be transitive, the pair $(x, z) = (11, 3)$ must be in $R$. Let's check if $(11, 3) \in R$. For this, we need $11 \in N, 3 \in N$ and $2(11) + 3 = 41$.

$2(11) + 3 = 22 + 3 = 25$.

Since $25 \neq 41$, the pair $(11, 3) \notin R$.

We found $(11, 19) \in R$ and $(19, 3) \in R$, but $(11, 3) \notin R$.

Therefore, the relation $R$ is not transitive.

Summary of Verification:

• $R$ is not reflexive.
• $R$ is not symmetric.
• $R$ is not transitive.

Given:

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

Set $B = \{2, 5, 6, 7\}$

(a) An injective mapping from A to B

An injective mapping (or one-to-one function) $f: A \to B$ is a function such that for any distinct elements $a_1, a_2 \in A$, their images $f(a_1), f(a_2)$ are distinct elements in $B$. Mathematically, if $a_1 \neq a_2$, then $f(a_1) \neq f(a_2)$. Equivalently, if $f(a_1) = f(a_2)$, then $a_1 = a_2$.

Since $|A| = 3$ and $|B| = 4$, it is possible to define an injective mapping from $A$ to $B$ because the number of elements in the domain ($A$) is less than or equal to the number of elements in the codomain ($B$).

Let's define a mapping $f: A \to B$ by assigning each element in $A$ to a unique element in $B$. For example:

Let $f(2) = 5$

Let $f(3) = 6$

Let $f(4) = 7$

We can represent this mapping as a set of ordered pairs:

$f = \{(2, 5), (3, 6), (4, 7)\}$.

In this mapping, each element in $A$ is mapped to a distinct element in $B$. For example, $2 \neq 3$, and $f(2) = 5 \neq f(3) = 6$. Similarly for all other pairs of distinct elements in $A$.

Therefore, $f$ is an injective mapping from $A$ to $B$.

(b) A mapping from A to B which is not injective

A mapping $f: A \to B$ is not injective if there exist at least two distinct elements $a_1, a_2 \in A$ such that $f(a_1) = f(a_2)$.

To construct such a mapping, we need to map two different elements from $A$ to the same element in $B$. For example, let's map the elements 2 and 3 from $A$ to the element 5 in $B$. The remaining element 4 from $A$ can be mapped to any element in $B$. Let's map it to 6.

Let's define a mapping $g: A \to B$ as follows:

Let $g(2) = 5$

Let $g(3) = 5$

Let $g(4) = 6$

We can represent this mapping as a set of ordered pairs:

$g = \{(2, 5), (3, 5), (4, 6)\}$.

This is a valid mapping from $A$ to $B$ because every element in $A$ is mapped to exactly one element in $B$.

This mapping is not injective because $2 \in A$ and $3 \in A$ are distinct elements ($2 \neq 3$), but their images are the same ($g(2) = 5$ and $g(3) = 5$).

Therefore, $g$ is a mapping from $A$ to $B$ which is not injective.

(c) A mapping from B to A

A mapping (or function) $h: B \to A$ is a rule that assigns to each element in the set $B$ (the domain) exactly one element in the set $A$ (the codomain).

The domain is $B = \{2, 5, 6, 7\}$, and the codomain is $A = \{2, 3, 4\}$. We need to define $h(y)$ for each $y \in B$, where $h(y)$ must be an element of $A$.

Let's define a mapping $h: B \to A$. For example:

Let $h(2) = 2$

Let $h(5) = 3$

Let $h(6) = 4$

Let $h(7) = 2$

We can represent this mapping as a set of ordered pairs:

$h = \{(2, 2), (5, 3), (6, 4), (7, 2)\}$.

This is a valid mapping from $B$ to $A$ because each element in $B$ is assigned to exactly one element in $A$. For example, the element 2 from $B$ is mapped to the element 2 in $A$; the element 5 from $B$ is mapped to the element 3 in $A$, and so on.

Note that since $|B| = 4$ and $|A| = 3$, a mapping from $B$ to $A$ cannot be injective because there are more elements in the domain than in the codomain (by the Pigeonhole Principle, at least two elements in $B$ must map to the same element in $A$). In our example $h$, the elements 2 and 7 from $B$ both map to the element 2 in $A$.

Therefore, $h$ is an example of a mapping from $B$ to $A$.

Let's provide examples of mappings based on injectivity (one-one) and surjectivity (onto).

(i) A map which is one-one but not onto

We need a function where every element in the domain maps to a unique element in the codomain, but the range is a proper subset of the codomain (i.e., not all elements in the codomain are mapped to).

Consider the sets:

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

Codomain $B = \{1, 2, 3, 4\}$

Let's define the function $f: A \to B$ as:

$f(x) = x$

In terms of ordered pairs, the function is:

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

Check for one-one:

For any distinct elements in $A$, their images in $B$ are distinct.

• $1 \neq 2$, $f(1) = 1 \neq f(2) = 2$.
• $1 \neq 3$, $f(1) = 1 \neq f(3) = 3$.
• $2 \neq 3$, $f(2) = 2 \neq f(3) = 3$.

Since $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in A$, the function $f$ is one-one.

Check for onto:

The function is onto if for every element $y$ in the codomain $B$, there exists an element $x$ in the domain $A$ such that $f(x) = y$.

The codomain is $B = \{1, 2, 3, 4\}$. The range of $f$ is $\{f(1), f(2), f(3)\} = \{1, 2, 3\}$.

The element $4 \in B$ is in the codomain, but there is no element $x \in A$ such that $f(x) = 4$.

Since the range $\{1, 2, 3\}$ is a proper subset of the codomain $\{1, 2, 3, 4\}$, the function $f$ is not onto.

Thus, $f: \{1, 2, 3\} \to \{1, 2, 3, 4\}$ defined by $f(x) = x$ is a map which is one-one but not onto.

An example with infinite sets: $f: N \to N$ defined by $f(x) = x+1$. The domain is $N = \{1, 2, 3, \dots\}$ and the codomain is $N = \{1, 2, 3, \dots\}$. This is one-one but the element 1 is not in the range.

(ii) A map which is not one-one but onto

We need a function where at least two distinct elements in the domain map to the same element in the codomain, but the range is equal to the codomain.

Consider the sets:

Domain $A = \{a, b, c, d\}$

Codomain $B = \{1, 2, 3\}$

Let's define the function $g: A \to B$ as:

$g(a) = 1$

$g(b) = 2$

$g(c) = 3$

$g(d) = 1$

In terms of ordered pairs, the function is:

$g = \{(a, 1), (b, 2), (c, 3), (d, 1)\}$.

Check for one-one:

The elements $a \in A$ and $d \in A$ are distinct ($a \neq d$), but their images are the same ($g(a) = 1$ and $g(d) = 1$).

Therefore, the function $g$ is not one-one.

Check for onto:

The codomain is $B = \{1, 2, 3\}$. The range of $g$ is $\{g(a), g(b), g(c), g(d)\} = \{1, 2, 3, 1\} = \{1, 2, 3\}$.

The range $\{1, 2, 3\}$ is equal to the codomain $\{1, 2, 3\}$.

For every element in $B$, there is at least one element in $A$ that maps to it (1 is mapped by $a$ and $d$, 2 by $b$, 3 by $c$).

Therefore, the function $g$ is onto.

Thus, $g: \{a, b, c, d\} \to \{1, 2, 3\}$ defined as above is a map which is not one-one but onto.

An example with infinite sets: $g: Z \to \{0, 1\}$ defined by $g(x) = x \pmod 2$. The domain is $Z = \{\dots, -1, 0, 1, \dots\}$ and the codomain is $\{0, 1\}$. This is not one-one (e.g., $g(0)=g(2)=0$) but onto (0 and 1 are both in the range).

(iii) A map which is neither one-one nor onto

We need a function where at least two distinct elements in the domain map to the same element in the codomain, and at least one element in the codomain is not the image of any element in the domain.

Consider the sets:

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

Codomain $B = \{a, b, c\}$

Let's define the function $h: A \to B$ as:

$h(1) = a$

$h(2) = a$

$h(3) = b$

In terms of ordered pairs, the function is:

$h = \{(1, a), (2, a), (3, b)\}$.

Check for one-one:

The elements $1 \in A$ and $2 \in A$ are distinct ($1 \neq 2$), but their images are the same ($h(1) = a$ and $h(2) = a$).

Therefore, the function $h$ is not one-one.

Check for onto:

The codomain is $B = \{a, b, c\}$. The range of $h$ is $\{h(1), h(2), h(3)\} = \{a, a, b\} = \{a, b\}$.

The element $c \in B$ is in the codomain, but there is no element $x \in A$ such that $h(x) = c$.

Since the range $\{a, b\}$ is a proper subset of the codomain $\{a, b, c\}$, the function $h$ is not onto.

Thus, $h: \{1, 2, 3\} \to \{a, b, c\}$ defined as above is a map which is neither one-one nor onto.

An example with infinite sets: $h: Z \to Z$ defined by $h(x) = x^2$. The domain is $Z$ and the codomain is $Z$. This is not one-one (e.g., $h(-1)=h(1)=1$) and not onto (e.g., $-1$ is not in the range).

Given:

Sets $A = \mathbb{R} \setminus \{3\}$ and $B = \mathbb{R} \setminus \{1\}$.

Function $f: A \to B$ defined by $f(x) = \frac{x - 2}{x - 3}$ for all $x \in A$.

To Show:

The function $f$ is bijective.

Solution:

To show that $f$ is bijective, we need to prove that it is both injective (one-one) and surjective (onto).

Injectivity (One-one):

Assume $f(x_1) = f(x_2)$ for arbitrary $x_1, x_2 \in A$. We must show that $x_1 = x_2$.

$f(x_1) = f(x_2)$

$\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}$

Since $x_1, x_2 \in A = \mathbb{R} \setminus \{3\}$, we have $x_1 - 3 \neq 0$ and $x_2 - 3 \neq 0$. We can cross-multiply.

$(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$

Expand both sides:

$x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6$

Subtract $x_1x_2$ from both sides:

$- 3x_1 - 2x_2 + 6 = - 3x_2 - 2x_1 + 6$

Subtract 6 from both sides:

$- 3x_1 - 2x_2 = - 3x_2 - 2x_1$

Add $3x_2$ to both sides:

$- 3x_1 + x_2 = - 2x_1$

Add $3x_1$ to both sides:

$x_2 = x_1$

Thus, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Therefore, $f$ is injective.

Surjectivity (Onto):

For any $y \in B$, we must find an $x \in A$ such that $f(x) = y$.

Let $y$ be an arbitrary element in $B = \mathbb{R} \setminus \{1\}$. We set $f(x) = y$ and solve for $x$ in terms of $y$.

$y = \frac{x - 2}{x - 3}$

Since $y \in B = \mathbb{R} \setminus \{1\}$, $y \neq 1$.

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

$yx - 3y = x - 2$

Rearrange the terms to isolate $x$:

$yx - x = 3y - 2$

Factor out $x$ from the left side:

$x(y - 1) = 3y - 2$

Since $y \neq 1$, $y - 1 \neq 0$, so we can divide by $(y - 1)$.

$x = \frac{3y - 2}{y - 1}$

We have found an expression for $x$. Now we need to check if this $x$ is in the domain $A = \mathbb{R} \setminus \{3\}$. That is, we must ensure $x \neq 3$ for any $y \in B$.

Suppose for the sake of contradiction that $x = 3$ for some $y \in B$.

$3 = \frac{3y - 2}{y - 1}$

$3(y - 1) = 3y - 2$

$3y - 3 = 3y - 2$

$-3 = -2$

This is a contradiction. Therefore, for any $y \in B$, the corresponding $x = \frac{3y - 2}{y - 1}$ cannot be equal to 3. Hence, $x \in \mathbb{R} \setminus \{3\}$, which is the domain $A$.

Thus, for every $y \in B$, there exists an $x \in A$ (namely $x = \frac{3y - 2}{y - 1}$) such that $f(x) = y$. Therefore, $f$ is surjective.

Conclusion:

Since the function $f$ is both injective and surjective, it is bijective.

Given:

Set $A = [-1, 1]$.

We are asked to discuss whether the following functions defined on $A$ (meaning domain is $A$ and codomain is $A$) are one-one, onto, or bijective:

• (i) $f(x) = \frac{x}{2}$
• (ii) $g(x) = |x|$
• (iii) $h(x) = x|x|$
• (iv) $k(x) = x^2$

To Discuss:

For each function, determine if it is one-one (injective), onto (surjective), or bijective (both one-one and onto) on the domain and codomain $A = [-1, 1]$.

Discussion:

Recall that for a function $\phi: A \to A$, where $A = [-1, 1]$:

• $\phi$ is one-one if for all $x_1, x_2 \in A$, $\phi(x_1) = \phi(x_2) \implies x_1 = x_2$.
• $\phi$ is onto if for every $y \in A$, there exists an $x \in A$ such that $\phi(x) = y$. This means the range of $\phi$ must be equal to the codomain $[-1, 1]$.
• $\phi$ is bijective if it is both one-one and onto.

(i) $f(x) = \frac{x}{2}$

Domain: $A = [-1, 1]$

Codomain: $A = [-1, 1]$

Injectivity (One-one):

Let $x_1, x_2 \in [-1, 1]$ such that $f(x_1) = f(x_2)$.

$\frac{x_1}{2} = \frac{x_2}{2}$

Multiplying both sides by 2, we get $x_1 = x_2$.

Since $f(x_1) = f(x_2) \implies x_1 = x_2$, the function $f$ is one-one.

Surjectivity (Onto):

Let $y$ be an arbitrary element in the codomain $[-1, 1]$. We need to find an $x \in [-1, 1]$ such that $f(x) = y$.

$f(x) = y$

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

$x = 2y$

Now we check if this value of $x$ is always in the domain $[-1, 1]$ for every $y \in [-1, 1]$.

If $y \in [-1, 1]$, then $-1 \le y \le 1$.

Multiplying the inequality by 2, we get $-2 \le 2y \le 2$.

So $x = 2y$ lies in the interval $[-2, 2]$. However, the domain is $[-1, 1]$.

For $x$ to be in the domain, we require $-1 \le x \le 1$, which means $-1 \le 2y \le 1$.

Dividing by 2, we get $-\frac{1}{2} \le y \le \frac{1}{2}$.

This shows that $f(x) = y$ has a solution $x$ in the domain $[-1, 1]$ only when $y$ is in the interval $[-\frac{1}{2}, \frac{1}{2}]$.

For example, if we choose $y=1 \in [-1, 1]$ (codomain), the required $x$ is $x = 2(1) = 2$. But $2 \notin [-1, 1]$ (domain).

The range of $f(x) = \frac{x}{2}$ for $x \in [-1, 1]$ is $f([-1, 1]) = [\frac{-1}{2}, \frac{1}{2}]$.

Since the range $[-\frac{1}{2}, \frac{1}{2}]$ is a proper subset of the codomain $[-1, 1]$, the function $f$ is not onto.

Bijectivity:

Since $f$ is one-one but not onto, it is not bijective.

(ii) $g(x) = |x|$

Domain: $A = [-1, 1]$

Codomain: $A = [-1, 1]$

The function is defined as $g(x) = |x| = \begin{cases} x & , & x \ge 0 \\ -x & , & x < 0 \end{cases}$.

Injectivity (One-one):

Let's check if distinct elements map to the same value.

Consider $x_1 = -0.5$ and $x_2 = 0.5$. Both $x_1, x_2 \in [-1, 1]$ and $x_1 \neq x_2$.

$g(x_1) = |-0.5| = 0.5$

$g(x_2) = |0.5| = 0.5$

Since $x_1 \neq x_2$ but $g(x_1) = g(x_2)$, the function $g$ is not one-one.

Surjectivity (Onto):

The range of $g(x) = |x|$ for $x \in [-1, 1]$ is the set of values $|x|$ can take. As $x$ varies from -1 to 1, $|x|$ varies from $|-1|=1$ down to $|0|=0$ and then up to $|1|=1$.

The range of $g$ is $[0, 1]$.

The codomain is $[-1, 1]$.

The range $[0, 1]$ is a proper subset of the codomain $[-1, 1]$. For example, the element $-0.1 \in [-1, 1]$ (codomain), but there is no $x \in [-1, 1]$ such that $g(x) = |x| = -0.1$ (since absolute values are always non-negative).

Thus, the function $g$ is not onto.

Bijectivity:

Since $g$ is neither one-one nor onto, it is not bijective.

(iii) $h(x) = x|x|$

Domain: $A = [-1, 1]$

Codomain: $A = [-1, 1]$

The function is defined as $h(x) = x|x| = \begin{cases} x \cdot x = x^2 & , & x \ge 0 \\ x \cdot (-x) = -x^2 & , & x < 0 \end{cases}$.

So, $h(x) = \begin{cases} x^2 & , & x \in [0, 1] \\ -x^2 & , & x \in [-1, 0) \end{cases}$.

Injectivity (One-one):

Let $x_1, x_2 \in [-1, 1]$ such that $h(x_1) = h(x_2)$.

Case 1: $x_1 \ge 0$ and $x_2 \ge 0$. Then $h(x_1) = x_1^2$ and $h(x_2) = x_2^2$. $x_1^2 = x_2^2$. Since $x_1, x_2 \ge 0$, $x_1 = x_2$.

Case 2: $x_1 < 0$ and $x_2 < 0$. Then $h(x_1) = -x_1^2$ and $h(x_2) = -x_2^2$. $-x_1^2 = -x_2^2 \implies x_1^2 = x_2^2$. Since $x_1, x_2 < 0$, $x_1 = -|x_1|$ and $x_2 = -|x_2|$. $x_1^2 = x_2^2 \implies |x_1| = |x_2|$. Since both are negative, $x_1 = x_2$.

Case 3: $x_1 \ge 0$ and $x_2 < 0$. Then $h(x_1) = x_1^2 \in [0, 1]$ and $h(x_2) = -x_2^2 \in [-1, 0)$. If $h(x_1) = h(x_2)$, then $x_1^2 = -x_2^2$. The left side is non-negative, and the right side is non-positive (since $x_2 \neq 0$, $-x_2^2 < 0$). The only way they can be equal is if both are 0, which implies $x_1^2 = 0$ and $-x_2^2 = 0$. This happens only if $x_1 = 0$ and $x_2 = 0$. But this contradicts $x_2 < 0$. So $h(x_1) = h(x_2)$ with $x_1 \ge 0$ and $x_2 < 0$ is only possible if $x_1=0$, $h(x_1)=0$, and $h(x_2)=0$, which implies $x_2=0$, contradiction. Thus, $h(x_1) = h(x_2)$ is only possible if $x_1$ and $x_2$ have the same sign (or both are 0).

From the cases, $h(x_1) = h(x_2) \implies x_1 = x_2$. The function $h$ is one-one.

Surjectivity (Onto):

Let $y$ be an arbitrary element in the codomain $[-1, 1]$. We need to find an $x \in [-1, 1]$ such that $h(x) = y$.

Case 1: $y \in [0, 1]$. We seek $x \in [-1, 1]$ such that $h(x) = y$. If $x \ge 0$, $h(x) = x^2 = y$. Since $y \ge 0$, $x = \sqrt{y}$. Since $y \in [0, 1]$, $\sqrt{y} \in [0, 1]$. This value of $x$ is in the domain $[-1, 1]$ and satisfies $x \ge 0$. So $x = \sqrt{y}$ is a valid preimage for $y$.

Case 2: $y \in [-1, 0)$. We seek $x \in [-1, 1]$ such that $h(x) = y$. If $x < 0$, $h(x) = -x^2 = y$. So $x^2 = -y$. Since $y < 0$, $-y > 0$. $x = \pm \sqrt{-y}$. Since we assume $x < 0$, we must choose $x = -\sqrt{-y}$. Since $y \in [-1, 0)$, $-y \in (0, 1]$. Then $\sqrt{-y} \in (0, 1]$. So $x = -\sqrt{-y} \in [-1, 0)$. This value of $x$ is in the domain $[-1, 1]$ and satisfies $x < 0$. So $x = -\sqrt{-y}$ is a valid preimage for $y$.

For every $y \in [-1, 1]$, we found an $x \in [-1, 1]$ such that $h(x) = y$. Specifically, $x = \sqrt{y}$ if $y \ge 0$, and $x = -\sqrt{-y}$ if $y < 0$.

The range of $h$ is the entire codomain $[-1, 1]$. Thus, the function $h$ is onto.

Bijectivity:

Since $h$ is both one-one and onto, it is bijective.

(iv) $k(x) = x^2$

Domain: $A = [-1, 1]$

Codomain: $A = [-1, 1]$

Injectivity (One-one):

Let's check if distinct elements map to the same value.

Consider $x_1 = -1$ and $x_2 = 1$. Both $x_1, x_2 \in [-1, 1]$ and $x_1 \neq x_2$.

$k(x_1) = (-1)^2 = 1$

$k(x_2) = (1)^2 = 1$

Since $x_1 \neq x_2$ but $k(x_1) = k(x_2)$, the function $k$ is not one-one.

(Another example: $x_1 = -0.5$, $x_2 = 0.5$. $k(-0.5) = (-0.5)^2 = 0.25$, $k(0.5) = (0.5)^2 = 0.25$.)

Surjectivity (Onto):

The range of $k(x) = x^2$ for $x \in [-1, 1]$ is the set of values $x^2$ can take. As $x$ varies from -1 to 1, $x^2$ varies from $(-1)^2=1$ down to $(0)^2=0$ and then up to $(1)^2=1$.

The range of $k$ is $[0, 1]$.

The codomain is $[-1, 1]$.

The range $[0, 1]$ is a proper subset of the codomain $[-1, 1]$. For example, the element $-0.1 \in [-1, 1]$ (codomain), but there is no $x \in [-1, 1]$ such that $k(x) = x^2 = -0.1$ (since squares of real numbers are always non-negative).

Thus, the function $k$ is not onto.

Bijectivity:

Since $k$ is neither one-one nor onto, it is not bijective.

Given:

Set $N$ of natural numbers, $N = \{1, 2, 3, \ldots\}$.

Four relations defined on $N$:

• (i) $R_1 = \{(x, y) : x \in N, y \in N, x > y\}$
• (ii) $R_2 = \{(x, y) : x \in N, y \in N, x + y = 10\}$
• (iii) $R_3 = \{(x, y) : x \in N, y \in N, xy \text{ is a square of an integer}\}$
• (iv) $R_4 = \{(x, y) : x \in N, y \in N, x + 4y = 10\}$

To Determine:

For each relation, verify if it is reflexive, symmetric, and transitive.

Analysis of Relations:

(i) Relation $R_1: x > y$ on $N$

Reflexivity:

A relation $R$ on $N$ is reflexive if $(a, a) \in R$ for all $a \in N$. For $R_1$, this means $a > a$ must be true for all $a \in N$. This is false for any natural number $a$, as a number cannot be strictly greater than itself. For example, for $a=1 \in N$, $(1, 1) \notin R_1$ because $1 \ngtr 1$.

Therefore, $R_1$ is not reflexive.

Symmetry:

A relation $R$ on $N$ is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$ for all $a, b \in N$. For $R_1$, this means if $a > b$, then $b > a$ must be true. This is false. For example, for $a=2, b=1 \in N$, $(2, 1) \in R_1$ because $2 > 1$. However, $(1, 2) \notin R_1$ because $1 \ngtr 2$.

Therefore, $R_1$ is not symmetric.

Transitivity:

A relation $R$ on $N$ is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in N$. For $R_1$, this means if $a > b$ and $b > c$, then $a > c$ must be true. This is a fundamental property of the "greater than" relation on numbers. If a is greater than b, and b is greater than c, then a must be greater than c. For example, if $a=3, b=2, c=1 \in N$, $(3, 2) \in R_1$ because $3 > 2$, and $(2, 1) \in R_1$ because $2 > 1$. The conclusion is $(3, 1)$, and indeed $(3, 1) \in R_1$ because $3 > 1$.

Therefore, $R_1$ is transitive.

(ii) Relation $R_2: x + y = 10$ on $N$

Reflexivity:

For $R_2$ to be reflexive, $(a, a) \in R_2$ for all $a \in N$. This means $a + a = 10$, or $2a = 10$, which implies $a = 5$. This is only true for the specific natural number $a=5$, not for all $a \in N$. For example, for $a=1 \in N$, $(1, 1) \notin R_2$ because $1 + 1 = 2 \neq 10$.

Therefore, $R_2$ is not reflexive.

Symmetry:

For $R_2$ to be symmetric, if $(a, b) \in R_2$, then $(b, a) \in R_2$. This means if $a + b = 10$, then $b + a = 10$ must be true. By the commutative property of addition, $a + b = b + a$, so if $a + b = 10$, then $b + a = 10$ is also true. For example, $(1, 9) \in R_2$ because $1 + 9 = 10$, and $(9, 1) \in R_2$ because $9 + 1 = 10$.

Therefore, $R_2$ is symmetric.

Transitivity:

For $R_2$ to be transitive, if $(a, b) \in R_2$ and $(b, c) \in R_2$, then $(a, c) \in R_2$. This means if $a + b = 10$ and $b + c = 10$, then $a + c = 10$ must be true. Let's consider a counterexample. We know $(1, 9) \in R_2$ since $1 + 9 = 10$, and $(9, 1) \in R_2$ since $9 + 1 = 10$. We have $(1, 9) \in R_2$ and $(9, 1) \in R_2$. According to the transitivity property, $(1, 1)$ must be in $R_2$. However, $1 + 1 = 2 \neq 10$, so $(1, 1) \notin R_2$.

Therefore, $R_2$ is not transitive.

(iii) Relation $R_3: xy$ is a square of an integer on $N$

Reflexivity:

For $R_3$ to be reflexive, $(a, a) \in R_3$ for all $a \in N$. This means $a \cdot a = a^2$ must be a square of an integer. For any natural number $a$, $a^2$ is the square of the integer $a$. So, $(a, a) \in R_3$ for all $a \in N$.

Therefore, $R_3$ is reflexive.

Symmetry:

For $R_3$ to be symmetric, if $(a, b) \in R_3$, then $(b, a) \in R_3$. This means if $ab$ is a square of an integer, then $ba$ must be a square of an integer. Since $ab = ba$, if $ab$ is a square, $ba$ is also the same square. For example, $(2, 8) \in R_3$ because $2 \cdot 8 = 16 = 4^2$, which is a square. Then $(8, 2) \in R_3$ because $8 \cdot 2 = 16 = 4^2$, which is a square.

Therefore, $R_3$ is symmetric.

Transitivity:

For $R_3$ to be transitive, if $(a, b) \in R_3$ and $(b, c) \in R_3$, then $(a, c) \in R_3$. This means if $ab = m^2$ and $bc = n^2$ for some integers $m, n$, then $ac$ must be a square of an integer. Consider the prime factorization of $a, b, c$. For $ab$ to be a square, the exponent of every prime in its prime factorization must be even. This means for every prime $p$, the sum of the exponents of $p$ in the prime factorizations of $a$ and $b$, denoted $v_p(a) + v_p(b)$, must be even. Similarly, $v_p(b) + v_p(c)$ must be even for all primes $p$. $v_p(a) + v_p(b)$ is even $\implies v_p(a)$ and $v_p(b)$ have the same parity. $v_p(b) + v_p(c)$ is even $\implies v_p(b)$ and $v_p(c)$ have the same parity. Combining these, $v_p(a)$, $v_p(b)$, and $v_p(c)$ all have the same parity for each prime $p$. Then $v_p(a) + v_p(c)$ must be even for all primes $p$. Since the exponent of every prime in the factorization of $ac$ is even, $ac$ is a perfect square. For example, $(1, 4) \in R_3$ ($1 \cdot 4 = 4 = 2^2$) and $(4, 9) \in R_3$ ($4 \cdot 9 = 36 = 6^2$). Check $(1, 9)$: $1 \cdot 9 = 9 = 3^2$, which is a square. So $(1, 9) \in R_3$.

Therefore, $R_3$ is transitive.

(iv) Relation $R_4: x + 4y = 10$ on $N$

Let's first find the pairs $(x, y) \in N \times N$ that satisfy $x + 4y = 10$. Since $x, y \in N$, $x \ge 1$ and $y \ge 1$. From $x + 4y = 10$, we get $x = 10 - 4y$. If $y=1$, $x = 10 - 4(1) = 6$. Since $6 \in N$, $(6, 1) \in R_4$. If $y=2$, $x = 10 - 4(2) = 2$. Since $2 \in N$, $(2, 2) \in R_4$. If $y=3$, $x = 10 - 4(3) = 10 - 12 = -2$. Since $-2 \notin N$, no pairs exist for $y \ge 3$. The relation $R_4$ is the set of ordered pairs $R_4 = \{(6, 1), (2, 2)\}$.

Reflexivity:

For $R_4$ to be reflexive, $(a, a) \in R_4$ for all $a \in N$. This means $a + 4a = 10$, or $5a = 10$, which implies $a = 2$. The pair $(2, 2)$ is in $R_4$, but this is only one element in $N$. For example, for $a=1 \in N$, $(1, 1) \notin R_4$ because $1 + 4(1) = 5 \neq 10$.

Therefore, $R_4$ is not reflexive.

Symmetry:

For $R_4$ to be symmetric, if $(a, b) \in R_4$, then $(b, a) \in R_4$. Consider the pair $(6, 1) \in R_4$. Is $(1, 6) \in R_4$? We check if $1 + 4(6) = 10$. $1 + 24 = 25 \neq 10$. So $(1, 6) \notin R_4$. Consider the pair $(2, 2) \in R_4$. Is $(2, 2) \in R_4$? Yes. Since $(6, 1) \in R_4$ but $(1, 6) \notin R_4$, the relation is not symmetric.

Therefore, $R_4$ is not symmetric.

Transitivity:

For $R_4$ to be transitive, if $(a, b) \in R_4$ and $(b, c) \in R_4$, then $(a, c) \in R_4$. The relation $R_4 = \{(6, 1), (2, 2)\}$. We check all possible chains $(a, b), (b, c) \in R_4$. Case 1: $(a, b) = (6, 1)$. So $b=1$. We need a pair $(1, c) \in R_4$. Looking at $R_4$, there is no pair that starts with 1. Thus, the premise "$(6, 1) \in R_4$ and $(1, c) \in R_4$" is false for any $c$. When the premise of an implication is false, the implication is true. So transitivity holds for chains starting with $(6, 1)$. Case 2: $(a, b) = (2, 2)$. So $b=2$. We need a pair $(2, c) \in R_4$. Looking at $R_4$, the only pair that starts with 2 is $(2, 2)$. So $c=2$. The chain is $(2, 2), (2, 2)$. We need to check if $(a, c) = (2, 2)$ is in $R_4$. Yes, $(2, 2) \in R_4$. Transitivity holds for this chain. Since the transitivity condition holds for all possible cases where the premise is true (or holds vacuously where the premise is false), the relation is transitive.

Therefore, $R_4$ is transitive.

Summary:

• Relation (i) $x > y$: Not Reflexive, Not Symmetric, Transitive.
• Relation (ii) $x + y = 10$: Not Reflexive, Symmetric, Not Transitive.
• Relation (iii) $xy$ is square: Reflexive, Symmetric, Transitive.
• Relation (iv) $x + 4y = 10$: Not Reflexive, Not Symmetric, Transitive.

Given:

Set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

Relation $R$ defined on $A \times A$ such that for $(a, b), (c, d) \in A \times A$, $(a, b) R (c, d)$ if and only if $a + d = b + c$.

To Prove:

$R$ is an equivalence relation on $A \times A$.

To Find:

The equivalence class $[(2, 5)]$.

Proof:

To prove that $R$ is an equivalence relation, we need to show that it is reflexive, symmetric, and transitive.

(i) Reflexivity:

A relation $R$ on a set $S$ is reflexive if $(x, x) \in R$ for all $x \in S$. In this case, the set is $A \times A$, and an element is an ordered pair $(a, b) \in A \times A$. We need to check if $(a, b) R (a, b)$ for all $(a, b) \in A \times A$.

According to the definition of $R$, $(a, b) R (a, b)$ if $a + b = b + a$.

By the commutative property of addition, $a + b = b + a$ is always true for any natural numbers $a, b \in A$.

Thus, $(a, b) R (a, b)$ for all $(a, b) \in A \times A$.

Therefore, $R$ is reflexive.

(ii) Symmetry:

A relation $R$ on a set $S$ is symmetric if whenever $(x, y) \in R$, then $(y, x) \in R$ for all $x, y \in S$. Here, $x = (a, b)$ and $y = (c, d)$, where $(a, b), (c, d) \in A \times A$. We need to check if $(a, b) R (c, d) \implies (c, d) R (a, b)$.

Assume $(a, b) R (c, d)$. By the definition of $R$, this means $a + d = b + c$.

We need to show that $(c, d) R (a, b)$, which means $c + b = d + a$.

Given $a + d = b + c$. By the commutative property of addition, this can be written as $d + a = c + b$, or $c + b = d + a$.

This is exactly the condition for $(c, d) R (a, b)$.

Thus, $(a, b) R (c, d) \implies (c, d) R (a, b)$ for all $(a, b), (c, d) \in A \times A$.

Therefore, $R$ is symmetric.

(iii) Transitivity:

A relation $R$ on a set $S$ is transitive if whenever $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$ for all $x, y, z \in S$. Here, $x = (a, b)$, $y = (c, d)$, and $z = (e, f)$, where $(a, b), (c, d), (e, f) \in A \times A$. We need to check if $(a, b) R (c, d)$ and $(c, d) R (e, f) \implies (a, b) R (e, f)$.

Assume $(a, b) R (c, d)$ and $(c, d) R (e, f)$.

From $(a, b) R (c, d)$, we have $a + d = b + c$. Rearranging this equation, we get $a - b = c - d$.

From $(c, d) R (e, f)$, we have $c + f = d + e$. Rearranging this equation, we get $c - d = e - f$.

Now we have $a - b = c - d$ and $c - d = e - f$. By the transitivity of equality, we can conclude that $a - b = e - f$.

Rearranging this equation, we get $a + f = b + e$.

This is exactly the condition for $(a, b) R (e, f)$.

Thus, $(a, b) R (c, d)$ and $(c, d) R (e, f) \implies (a, b) R (e, f)$ for all $(a, b), (c, d), (e, f) \in A \times A$.

Therefore, $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation on $A \times A$.

Finding the equivalent class [(2, 5)]:

The equivalence class of an element $(x, y) \in A \times A$, denoted by $[(x, y)]$, is the set of all elements $(a, b) \in A \times A$ such that $(x, y) R (a, b)$.

For the element $(2, 5)$, the equivalence class $[(2, 5)]$ is the set of all $(a, b) \in A \times A$ such that $(2, 5) R (a, b)$.

According to the definition of $R$, $(2, 5) R (a, b)$ if $2 + b = 5 + a$.

We need to find all pairs $(a, b)$ such that $a \in \{1, 2, \ldots, 9\}$, $b \in \{1, 2, \ldots, 9\}$, and $2 + b = 5 + a$.

Rearranging the equation, we get $b - a = 5 - 2$, which simplifies to $b - a = 3$, or $b = a + 3$.

Now we list the pairs $(a, b)$ that satisfy this condition, keeping in mind that both $a$ and $b$ must be in the set $A = \{1, 2, \ldots, 9\}$.

• If $a = 1$, then $b = 1 + 3 = 4$. $(1, 4) \in A \times A$.
• If $a = 2$, then $b = 2 + 3 = 5$. $(2, 5) \in A \times A$.
• If $a = 3$, then $b = 3 + 3 = 6$. $(3, 6) \in A \times A$.
• If $a = 4$, then $b = 4 + 3 = 7$. $(4, 7) \in A \times A$.
• If $a = 5$, then $b = 5 + 3 = 8$. $(5, 8) \in A \times A$.
• If $a = 6$, then $b = 6 + 3 = 9$. $(6, 9) \in A \times A$.
• If $a = 7$, then $b = 7 + 3 = 10$. Since $10 \notin A$, the pair $(7, 10)$ is not in $A \times A$. We cannot take values of $a \ge 7$.

The possible values for $a$ are $1, 2, 3, 4, 5, 6$. The corresponding pairs $(a, b)$ in $A \times A$ satisfying $b = a + 3$ are $(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)$.

The equivalence class $[(2, 5)]$ is the set of these pairs.

The equivalent class $[(2, 5)] = \{(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)\}$.

Given:

A function $f: A \to B$.

To Prove:

The function $f$ is invertible if and only if $f$ is both one-one (injective) and onto (surjective).

Proof:

The statement "if and only if" requires us to prove two implications:

1. If $f$ is invertible, then $f$ is one-one and onto.
2. If $f$ is one-one and onto, then $f$ is invertible.

Part 1: If $f$ is invertible, then $f$ is one-one and onto.

Assume that $f: A \to B$ is invertible. By the definition of an invertible function, there exists a function $g: B \to A$ such that the composite functions satisfy:

$g \circ f = I_A$ (the identity function on set $A$)

$f \circ g = I_B$ (the identity function on set $B$)

where $I_A(a) = a$ for all $a \in A$, and $I_B(b) = b$ for all $b \in B$.

To show $f$ is one-one:

Let $a_1, a_2$ be arbitrary elements in $A$. Assume $f(a_1) = f(a_2)$.

Apply the function $g$ to both sides of the equation:

$g(f(a_1)) = g(f(a_2))$

By the definition of the composition of functions, this is $(g \circ f)(a_1) = (g \circ f)(a_2)$.

Since $g \circ f = I_A$, we have $I_A(a_1) = I_A(a_2)$.

By the definition of the identity function $I_A$, $a_1 = a_2$.

Thus, we have shown that $f(a_1) = f(a_2) \implies a_1 = a_2$. Therefore, $f$ is one-one.

To show $f$ is onto:

Let $b$ be an arbitrary element in the codomain $B$. We need to show that there exists at least one element $a \in A$ such that $f(a) = b.

Consider the element $g(b)$ in the set $A$ (since the domain of $g$ is $B$ and its codomain is $A$). Let $a = g(b)$.

Now, apply the function $f$ to this element $a$:

$f(a) = f(g(b))$

By the definition of the composition of functions, this is $(f \circ g)(b)$.

Since $f \circ g = I_B$, we have $(f \circ g)(b) = I_B(b)$.

By the definition of the identity function $I_B$, $I_B(b) = b$.

So, we have $f(a) = b$, where $a = g(b) \in A$.

Thus, for every $b \in B$, there exists an element $a \in A$ such that $f(a) = b$. Therefore, $f$ is onto.

We have successfully shown that if $f$ is invertible, then $f$ is both one-one and onto.

Part 2: If $f$ is one-one and onto, then $f$ is invertible.

Assume that $f: A \to B$ is both one-one and onto. We need to show that $f$ is invertible by constructing an inverse function $g: B \to A$ such that $g \circ f = I_A$ and $f \circ g = I_B$.

Since $f$ is onto, for every element $b \in B$, there exists at least one element $a \in A$ such that $f(a) = b$.

Since $f$ is one-one, this element $a$ in $A$ is unique. Suppose there were two elements $a_1, a_2 \in A$ such that $f(a_1) = b$ and $f(a_2) = b$. Then $f(a_1) = f(a_2)$. Since $f$ is one-one, this implies $a_1 = a_2$. Thus, for each $b \in B$, there is exactly one element $a \in A$ such that $f(a) = b$.

Now we can define a function $g: B \to A$. For each $b \in B$, define $g(b)$ to be the unique element $a \in A$ such that $f(a) = b$. This defines a unique value $g(b)$ for every element $b$ in the domain $B$, so $g$ is a well-defined function from $B$ to $A$.

To show $g \circ f = I_A$:

Let $a$ be an arbitrary element in $A$. We want to evaluate $(g \circ f)(a) = g(f(a))$. Let $b = f(a)$. By the definition of $g$, $g(b)$ is the unique element $a'$ in $A$ such that $f(a') = b$. Since we know $f(a) = b$, and $a$ is in $A$, $a$ is that unique element. Therefore, $g(b) = a$. Substituting back $b = f(a)$, we get $g(f(a)) = a$. This holds for all $a \in A$. Thus, $g \circ f = I_A$.

To show $f \circ g = I_B$:

Let $b$ be an arbitrary element in $B$. We want to evaluate $(f \circ g)(b) = f(g(b))$. By the definition of $g$, $g(b)$ is the unique element $a$ in $A$ such that $f(a) = b$. So, if we let $a = g(b)$, then by definition, $f(a) = b$. Substituting back $a = g(b)$, we get $f(g(b)) = b$. This holds for all $b \in B$. Thus, $f \circ g = I_B$.

We have constructed a function $g: B \to A$ such that $g \circ f = I_A$ and $f \circ g = I_B$. By the definition of an invertible function, this means $f$ is invertible, and $g$ is its inverse ($g = f^{-1}$).

Conclusion:

From Part 1, we proved that if $f$ is invertible, then $f$ is one-one and onto.

From Part 2, we proved that if $f$ is one-one and onto, then $f$ is invertible.

Therefore, a function $f: A \to B$ is invertible if and only if $f$ is both one-one and onto. This completes the proof.

Given:

Functions $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 + 3x + 1$.

Functions $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = 2x - 3$.

To Find:

The composite functions (i) $f \circ g$, (ii) $g \circ f$, (iii) $f \circ f$, and (iv) $g \circ g$.

Solution:

(i) $f \circ g$

The composite function $f \circ g$ is defined as $(f \circ g)(x) = f(g(x))$.

Substitute the expression for $g(x)$ into the function $f(x)$: $f(g(x)) = f(2x - 3)$

Replace every instance of $x$ in $f(x) = x^2 + 3x + 1$ with $(2x - 3)$: $f(2x - 3) = (2x - 3)^2 + 3(2x - 3) + 1$

Expand the terms:

$(2x - 3)^2 = (2x)^2 - 2(2x)(3) + (-3)^2 = 4x^2 - 12x + 9$

$3(2x - 3) = 6x - 9$

So, $f(2x - 3) = (4x^2 - 12x + 9) + (6x - 9) + 1$

Combine like terms: $f(2x - 3) = 4x^2 + (-12x + 6x) + (9 - 9 + 1)$ $f(2x - 3) = 4x^2 - 6x + 1$

Thus, $(f \circ g)(x) = 4x^2 - 6x + 1$.

(ii) $g \circ f$

The composite function $g \circ f$ is defined as $(g \circ f)(x) = g(f(x))$.

Substitute the expression for $f(x)$ into the function $g(x)$: $g(f(x)) = g(x^2 + 3x + 1)$

Replace every instance of $x$ in $g(x) = 2x - 3$ with $(x^2 + 3x + 1)$: $g(x^2 + 3x + 1) = 2(x^2 + 3x + 1) - 3$

Distribute the 2: $g(x^2 + 3x + 1) = 2x^2 + 6x + 2 - 3$

Combine the constant terms: $g(x^2 + 3x + 1) = 2x^2 + 6x - 1$

Thus, $(g \circ f)(x) = 2x^2 + 6x - 1$.

(iii) $f \circ f$

The composite function $f \circ f$ is defined as $(f \circ f)(x) = f(f(x))$.

Substitute the expression for $f(x)$ into the function $f(x)$: $f(f(x)) = f(x^2 + 3x + 1)$

Replace every instance of $x$ in $f(x) = x^2 + 3x + 1$ with $(x^2 + 3x + 1)$: $f(x^2 + 3x + 1) = (x^2 + 3x + 1)^2 + 3(x^2 + 3x + 1) + 1$

Expand the terms:

$(x^2 + 3x + 1)^2 = (x^2 + (3x + 1))^2 = (x^2)^2 + 2x^2(3x + 1) + (3x + 1)^2$ $= x^4 + 6x^3 + 2x^2 + (9x^2 + 6x + 1)$ $= x^4 + 6x^3 + 11x^2 + 6x + 1$

$3(x^2 + 3x + 1) = 3x^2 + 9x + 3$

So, $f(x^2 + 3x + 1) = (x^4 + 6x^3 + 11x^2 + 6x + 1) + (3x^2 + 9x + 3) + 1$

Combine like terms: $f(x^2 + 3x + 1) = x^4 + 6x^3 + (11x^2 + 3x^2) + (6x + 9x) + (1 + 3 + 1)$ $f(x^2 + 3x + 1) = x^4 + 6x^3 + 14x^2 + 15x + 5$

Thus, $(f \circ f)(x) = x^4 + 6x^3 + 14x^2 + 15x + 5$.

(iv) $g \circ g$

The composite function $g \circ g$ is defined as $(g \circ g)(x) = g(g(x))$.

Substitute the expression for $g(x)$ into the function $g(x)$: $g(g(x)) = g(2x - 3)$

Replace every instance of $x$ in $g(x) = 2x - 3$ with $(2x - 3)$: $g(2x - 3) = 2(2x - 3) - 3$

Distribute the 2: $g(2x - 3) = 4x - 6 - 3$

Combine the constant terms: $g(2x - 3) = 4x - 9$

Thus, $(g \circ g)(x) = 4x - 9$.

Given:

Set $Q$ of rational numbers.

Several binary operations $*$ defined on $Q$.

To Find:

Which of the given binary operations are commutative.

Definition:

A binary operation $*$ on a set $S$ is said to be commutative if $a * b = b * a$ for all $a, b \in S$. Here $S = Q$.

(i) $a * b = a - b$ for all $a, b \in Q$

We need to check if $a * b = b * a$, which means $a - b = b - a$ for all $a, b \in Q$.

Let's choose specific rational numbers, say $a=1$ and $b=2$.

$a * b = 1 - 2 = -1$

$b * a = 2 - 1 = 1$

Since $-1 \neq 1$, $a - b \neq b - a$ for all $a, b \in Q$.

Therefore, the binary operation $a * b = a - b$ is not commutative.

(ii) $a * b = a^2 + b^2$ for all $a, b \in Q$

We need to check if $a * b = b * a$, which means $a^2 + b^2 = b^2 + a^2$ for all $a, b \in Q$.

By the commutative property of addition of rational numbers, $a^2 + b^2$ is always equal to $b^2 + a^2$.

For example, if $a=\frac{1}{2}$ and $b=\frac{3}{4}$,

$a * b = (\frac{1}{2})^2 + (\frac{3}{4})^2 = \frac{1}{4} + \frac{9}{16} = \frac{4+9}{16} = \frac{13}{16}$

$b * a = (\frac{3}{4})^2 + (\frac{1}{2})^2 = \frac{9}{16} + \frac{1}{4} = \frac{9+4}{16} = \frac{13}{16}$

Since $a * b = b * a$ for all $a, b \in Q$, the binary operation $a * b = a^2 + b^2$ is commutative.

(iii) $a * b = a + ab$ for all $a, b \in Q$

We need to check if $a * b = b * a$, which means $a + ab = b + ba$ for all $a, b \in Q$.

Let's choose specific rational numbers, say $a=1$ and $b=2$.

$a * b = 1 + 1 \cdot 2 = 1 + 2 = 3$

$b * a = 2 + 2 \cdot 1 = 2 + 2 = 4$

Since $3 \neq 4$, $a + ab \neq b + ba$ for all $a, b \in Q$.

Therefore, the binary operation $a * b = a + ab$ is not commutative.

(iv) $a * b = (a - b)^2$ for all $a, b \in Q$

We need to check if $a * b = b * a$, which means $(a - b)^2 = (b - a)^2$ for all $a, b \in Q$.

We know that $(a - b)^2 = (-1 \cdot (b - a))^2 = (-1)^2 \cdot (b - a)^2 = 1 \cdot (b - a)^2 = (b - a)^2$.

Alternatively, expanding both sides:

$(a - b)^2 = a^2 - 2ab + b^2$

$(b - a)^2 = b^2 - 2ba + a^2$

By the commutative property of multiplication and addition of rational numbers, $ab = ba$ and $a^2 - 2ab + b^2 = a^2 + b^2 - 2ab = b^2 + a^2 - 2ba = (b - a)^2$.

Since $(a - b)^2 = (b - a)^2$ for all $a, b \in Q$, the binary operation $a * b = (a - b)^2$ is commutative.

Given:

A binary operation $*$ defined on $\mathbb{R}$ by $a * b = 1 + ab$ for all $a, b \in \mathbb{R}$.

To Determine:

Whether the operation $*$ is commutative, associative, both, or neither.

Analysis:

Commutativity:

A binary operation $*$ on $\mathbb{R}$ is commutative if $a * b = b * a$ for all $a, b \in \mathbb{R}$.

Left side: $a * b = 1 + ab$

Right side: $b * a = 1 + ba$

Since the multiplication of real numbers is commutative, $ab = ba$.

Therefore, $1 + ab = 1 + ba$, which means $a * b = b * a$ for all $a, b \in \mathbb{R}$.

The operation $*$ is commutative.

Associativity:

A binary operation $*$ on $\mathbb{R}$ is associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{R}$.

Let's evaluate the left side:

$(a * b) * c = (1 + ab) * c$

Using the definition of $*$, we replace the first operand $(1 + ab)$ with $x$ and the second operand $c$ with $y$ in $x * y = 1 + xy$:

$(1 + ab) * c = 1 + (1 + ab)c = 1 + c + abc$

Now, let's evaluate the right side:

$a * (b * c) = a * (1 + bc)$

Using the definition of $*$, we replace the first operand $a$ with $x$ and the second operand $(1 + bc)$ with $y$ in $x * y = 1 + xy$:

$a * (1 + bc) = 1 + a(1 + bc) = 1 + a + abc$

For the operation to be associative, we must have $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{R}$.

This requires $1 + c + abc = 1 + a + abc$ for all $a, b, c \in \mathbb{R}$.

Subtracting $1 + abc$ from both sides gives $c = a$.

This equation $c = a$ is not true for all $a, c \in \mathbb{R}$. For example, if $a=1$ and $c=2$, then $1 \neq 2$.

Let's choose specific values for $a, b, c$ to demonstrate non-associativity:

Let $a = 1, b = 2, c = 3$.

$(a * b) * c = (1 * 2) * 3 = (1 + 1 \cdot 2) * 3 = (1 + 2) * 3 = 3 * 3 = 1 + 3 \cdot 3 = 1 + 9 = 10$.

$a * (b * c) = 1 * (2 * 3) = 1 * (1 + 2 \cdot 3) = 1 * (1 + 6) = 1 * 7 = 1 + 1 \cdot 7 = 1 + 7 = 8$.

Since $10 \neq 8$, the operation $*$ is not associative.

Conclusion:

The operation $*$ is commutative but not associative.

Comparing with the given options:

(i) commutative but not associative - Matches our conclusion.

(ii) associative but not commutative - Incorrect.

(iii) neither commutative nor associative - Incorrect.

(iv) both commutative and associative - Incorrect.

The correct option is (i) commutative but not associative.

Solution:

The given relation R on the set T of all triangles is defined as $aRb$ if triangle $a$ is congruent to triangle $b$, for all $a, b \in T$. We need to check if R is reflexive, symmetric, and transitive.

Checking for Reflexivity:

A relation R is reflexive if $aRa$ for all $a \in T$.

For any triangle $a \in T$, triangle $a$ is congruent to itself.

$aRa$ holds true for all $a \in T$.

Therefore, R is reflexive.

Checking for Symmetry:

A relation R is symmetric if whenever $aRb$ holds, $bRa$ also holds for all $a, b \in T$.

Suppose $aRb$ holds. This means triangle $a$ is congruent to triangle $b$.

If triangle $a$ is congruent to triangle $b$, then by definition of congruence, triangle $b$ is also congruent to triangle $a$.

So, $bRa$ holds.

Thus, if $aRb$, then $bRa$.

Therefore, R is symmetric.

Checking for Transitivity:

A relation R is transitive if whenever $aRb$ and $bRc$ hold, $aRc$ also holds for all $a, b, c \in T$.

Suppose $aRb$ and $bRc$ hold.

This means triangle $a$ is congruent to triangle $b$, and triangle $b$ is congruent to triangle $c$.

If triangle $a$ is congruent to triangle $b$, and triangle $b$ is congruent to triangle $c$, then triangle $a$ must be congruent to triangle $c$. This is a property of congruence.

So, $aRc$ holds.

Thus, if $aRb$ and $bRc$, then $aRc$.

Therefore, R is transitive.

Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

The correct option is (C).

Solution:

Let S be the non-empty set consisting of children in a family. The relation R on S is defined as $aRb$ if $a$ is brother of $b$, for all $a, b \in S$.

Checking for Reflexivity:

A relation R is reflexive if $aRa$ for all $a \in S$.

For $aRa$ to hold, $a$ must be a brother of $a$. A child cannot be their own brother.

Therefore, $aRa$ does not hold for any $a \in S$.

R is not reflexive.

Checking for Symmetry:

A relation R is symmetric if whenever $aRb$ holds, $bRa$ also holds for all $a, b \in S$.

Suppose $aRb$ holds. This means $a$ is brother of $b$. This implies $a$ is male.

For $bRa$ to hold, $b$ must be brother of $a$. This implies $b$ must be male.

Consider the case where $a$ is a brother of $b$, and $b$ is a sister. Then $aRb$ holds, but $bRa$ does not hold because $b$ is not a brother ( $b$ is female).

Since we found a case where $aRb$ holds but $bRa$ does not, the relation is not symmetric.

R is not symmetric.

Checking for Transitivity:

A relation R is transitive if whenever $aRb$ and $bRc$ hold, $aRc$ also holds for all $a, b, c \in S$.

Suppose $aRb$ and $bRc$ hold.

$aRb$ means $a$ is brother of $b$. This implies $a$ is male and $a$ and $b$ are siblings.

$bRc$ means $b$ is brother of $c$. This implies $b$ is male and $b$ and $c$ are siblings.

From "$a$ and $b$ are siblings" and "$b$ and $c$ are siblings", it follows that $a, b,$ and $c$ are all siblings in the same family.

We know that $a$ is male (from $aRb$).

We need to check if $aRc$ holds, i.e., if $a$ is brother of $c$. Since $a$ is male and $a$ and $c$ are siblings, $a$ is indeed a brother of $c$.

Thus, if $aRb$ and $bRc$, then $aRc$.

R is transitive.

Based on the checks, the relation R is not reflexive, not symmetric, but transitive.

The correct option that matches this description is (B) transitive but not symmetric.

Solution:

An equivalence relation on a set A is a relation R that is reflexive, symmetric, and transitive.

There is a one-to-one correspondence between the set of equivalence relations on a set A and the set of partitions of A.

A partition of a set A is a collection of non-empty disjoint subsets of A whose union is A.

We need to find the number of partitions of the set $A = \{1, 2, 3\}$. The possible partitions are:

1. All elements in separate subsets:

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

2. Two elements in one subset, one element separate:

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

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

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

3. All elements in one subset:

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

Listing all the partitions:

Partition 1: $\{\{1\}, \{2\}, \{3\}\}$

Partition 2: $\{\{1, 2\}, \{3\}\}$

Partition 3: $\{\{1, 3\}, \{2\}\}$

Partition 4: $\{\{2, 3\}, \{1\}\}$

Partition 5: $\{\{1, 2, 3\}\}$

There are a total of 5 distinct partitions of the set $A = \{1, 2, 3\}$.

Each partition corresponds to exactly one equivalence relation on A.

Therefore, the maximum number of equivalence relations on the set A is 5.

The number of partitions of a set of size $n$ is given by the Bell number $B_n$. For $n=3$, $B_3 = 5$.

The correct option is (D).

Solution:

Let the set be $A = \{1, 2, 3\}$. The relation R on A is defined as $R = \{(1, 2)\}$.

Checking for Reflexivity:

A relation R on set A is reflexive if $(x, x) \in R$ for all $x \in A$.

For R to be reflexive, the pairs $(1, 1), (2, 2), (3, 3)$ must be in R.

However, $R = \{(1, 2)\}$, which does not contain $(1, 1)$, $(2, 2)$, or $(3, 3)$.

Therefore, R is not reflexive.

Checking for Symmetry:

A relation R on set A is symmetric if whenever $(x, y) \in R$, then $(y, x) \in R$ for all $x, y \in A$.

We have the pair $(1, 2) \in R$. For R to be symmetric, the pair $(2, 1)$ must also be in R.

However, $(2, 1) \notin R$ since $R = \{(1, 2)\}$.

Therefore, R is not symmetric.

Checking for Transitivity:

A relation R on set A is transitive if whenever $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$ for all $x, y, z \in A$.

We examine the pairs in R. The only pair is $(1, 2)$. Let $x = 1$ and $y = 2$. So $(x, y) = (1, 2) \in R$.

Now we need to check if there is any pair $(y, z) \in R$, i.e., a pair starting with $y=2$.

There is no pair in $R = \{(1, 2)\}$ that starts with 2.

The condition for transitivity "if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$" becomes "if (something false), then (something)". An implication with a false premise is always true.

Since there are no two pairs $(x, y) \in R$ and $(y, z) \in R$ where the second element of the first pair is the same as the first element of the second pair, the condition for transitivity is vacuously satisfied.

Therefore, R is transitive.

Based on the checks, the relation R is transitive, but neither reflexive nor symmetric.

The correct option is (B).

Solution:

Let the set be $\mathbb{R}$, the set of real numbers. The relation R on $\mathbb{R}$ is defined as $aRb$ if $a \geq b$, for all $a, b \in \mathbb{R}$.

Checking for Reflexivity:

A relation R is reflexive if $aRa$ for all $a \in \mathbb{R}$.

$aRa$ means $a \geq a$.

For any real number $a$, $a \geq a$ is always true.

Therefore, R is reflexive.

Checking for Symmetry:

A relation R is symmetric if whenever $aRb$ holds, $bRa$ also holds for all $a, b \in \mathbb{R}$.

Suppose $aRb$ holds. This means $a \geq b$.

For R to be symmetric, $bRa$ must hold, which means $b \geq a$.

Consider $a=3$ and $b=2$. We have $3 \geq 2$, so $3R2$ holds.

However, $2 \geq 3$ is false, so $2R3$ does not hold.

Since we found a case where $aRb$ holds but $bRa$ does not, the relation is not symmetric.

Therefore, R is not symmetric.

Checking for Transitivity:

A relation R is transitive if whenever $aRb$ and $bRc$ hold, $aRc$ also holds for all $a, b, c \in \mathbb{R}$.

Suppose $aRb$ and $bRc$ hold.

$aRb$ means $a \geq b$.

$bRc$ means $b \geq c$.

If $a \geq b$ and $b \geq c$, it follows that $a \geq c$. This is a fundamental property of inequalities.

So, $aRc$ holds.

Thus, if $aRb$ and $bRc$, then $aRc$.

Therefore, R is transitive.

Based on the checks, the relation R is reflexive and transitive, but not symmetric.

This matches option (B).

Solution:

The given set is $A = \{1, 2, 3\}$.

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

Checking for Reflexivity:

A relation R on set A is reflexive if $(x, x) \in R$ for all $x \in A$.

The elements in set A are 1, 2, and 3.

We need to check if $(1, 1) \in R$, $(2, 2) \in R$, and $(3, 3) \in R$.

From the definition of R, we see that $(1, 1) \in R$, $(2, 2) \in R$, and $(3, 3) \in R$.

Since $(x, x) \in R$ for all $x \in A$, the relation R is reflexive.

Checking for Symmetry:

A relation R on set A is symmetric if whenever $(x, y) \in R$, then $(y, x) \in R$ for all $x, y \in A$.

We need to check this condition for every pair in R.

Consider the pair $(1, 2) \in R$. For R to be symmetric, $(2, 1)$ must also be in R. However, $(2, 1) \notin R$.

Since there exists a pair $(1, 2) \in R$ such that $(2, 1) \notin R$, the relation R is not symmetric.

Checking for Transitivity:

A relation R on set A is transitive if whenever $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$ for all $x, y, z \in A$.

Let's check all possible combinations of pairs $(x, y) \in R$ and $(y, z) \in R$:

• $(1, 1) \in R$ and $(1, 2) \in R$. Check $(1, 2) \in R$. Yes.
• $(1, 1) \in R$ and $(1, 3) \in R$. Check $(1, 3) \in R$. Yes.
• $(1, 2) \in R$ and $(2, 2) \in R$. Check $(1, 2) \in R$. Yes.
• $(1, 2) \in R$ and $(2, 3) \in R$. Check $(1, 3) \in R$. Yes, $(1, 3) \in R$.
• $(1, 3) \in R$. There is no pair in R starting with 3 other than (3,3). $(1, 3) \in R$ and $(3, 3) \in R$. Check $(1, 3) \in R$. Yes.
• $(2, 2) \in R$ and $(2, 3) \in R$. Check $(2, 3) \in R$. Yes.
• $(2, 3) \in R$. There is no pair in R starting with 3 other than (3,3). $(2, 3) \in R$ and $(3, 3) \in R$. Check $(2, 3) \in R$. Yes.
• $(3, 3) \in R$. There are no pairs in R starting with the second element 3 other than (3,3).

In all cases where $(x, y) \in R$ and $(y, z) \in R$, the pair $(x, z)$ is also in R.

Therefore, the relation R is transitive.

Summary: The relation R is reflexive, not symmetric, and transitive.

Comparing this with the given options, option (A) states that R is reflexive but not symmetric.

The correct option is (A).

Solution:

Let the set be $Q \setminus \{0\}$, the set of all non-zero rational numbers.

The binary operation * is defined as $a * b = \frac{ab}{2}$ for all $a, b \in Q \setminus \{0\}$.

An element $e \in Q \setminus \{0\}$ is called the identity element for the operation * if for all $a \in Q \setminus \{0\}$, we have:

$a * e = a$ and $e * a = a$.

Consider the condition $a * e = a$.

Using the definition of the operation, we have:

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

Since $a \in Q \setminus \{0\}$, $a$ is a non-zero rational number. We can multiply both sides by 2 and divide both sides by $a$ (since $a \neq 0$).

$ae = 2a$

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

$e = 2$

Now, consider the condition $e * a = a$.

Using the definition of the operation, we have:

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

Multiplying both sides by 2 and dividing by $a$ (since $a \neq 0$):

$ea = 2a$

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

$e = 2$

Both conditions yield the same value for the identity element, $e=2$.

We must check if $e = 2$ belongs to the set $Q \setminus \{0\}$. Since 2 is a non-zero rational number, $2 \in Q \setminus \{0\}$.

Thus, the identity element for the binary operation * on $Q \setminus \{0\}$ is 2.

The correct option is (C).

Solution:

Let the set A have 5 elements, so $|A| = 5$.

Let the set B have 6 elements, so $|B| = 6$.

We are looking for the number of one-one (injective) and onto (surjective) mappings from set A to set B.

A mapping that is both one-one and onto is called a bijective mapping or a bijection.

For a bijection to exist between two finite sets A and B, it is necessary that the two sets have the same number of elements.

That is, a bijection $f: A \to B$ exists if and only if $|A| = |B|$.

In this problem, we have $|A| = 5$ and $|B| = 6$.

Since $|A| \neq |B|$, there cannot be a bijection from set A to set B.

Specifically, since $|A| < |B|$, it is impossible to map the elements of A to cover all elements of B (i.e., the function cannot be onto/surjective).

Therefore, the number of one-one and onto mappings from A to B is 0.

The correct option is (C).

Solution:

Let the set A be $A = \{1, 2, 3, ..., n\}$, so the number of elements in A is $|A| = n$.

Let the set B be $B = \{a, b\}$, so the number of elements in B is $|B| = 2$.

A function $f: A \to B$ is said to be surjective (or onto) if for every element $y$ in the codomain B, there exists at least one element $x$ in the domain A such that $f(x) = y$.

In this case, a function $f: A \to \{a, b\}$ is surjective if its range is exactly $\{a, b\}$. This means that at least one element from A maps to $a$, and at least one element from A maps to $b$.

First, let's find the total number of functions from A to B.

For each element in A, there are two possible images in B (either $a$ or $b$). Since there are $n$ elements in A, the total number of functions from A to B is $2 \times 2 \times ... \times 2$ (n times), which is $2^n$.

Total number of functions from A to B = $2^n$.

Now, we need to find the number of functions that are not surjective.

A function $f: A \to B$ is not surjective if its range is a proper subset of B.

The proper subsets of $B = \{a, b\}$ are $\emptyset$, $\{a\}$, and $\{b\}$. However, the range of a function from a non-empty set cannot be empty. Since $A = \{1, 2, ..., n\}$, we assume $n \ge 1$ (as the set is described with elements $1, 2, ...$). If $n \ge 1$, A is non-empty, and the range must contain at least one element.

So, for $n \ge 1$, a function is not surjective if its range is either $\{a\}$ or $\{b\}$.

Case 1: The range of the function is $\{a\}$.

This means every element in A is mapped to the element $a$ in B. There is only one such function: $f(x) = a$ for all $x \in A$.

Number of functions whose range is $\{a\}$ = 1.

Case 2: The range of the function is $\{b\}$.

This means every element in A is mapped to the element $b$ in B. There is only one such function: $f(x) = b$ for all $x \in A$.

Number of functions whose range is $\{b\}$ = 1.

These two cases (range is $\{a\}$ and range is $\{b\}$) are mutually exclusive if $n \ge 1$, because a function cannot map all elements to $a$ and simultaneously map all elements to $b$ if the domain is non-empty.

So, the number of non-surjective functions is the sum of the numbers from Case 1 and Case 2.

Number of non-surjective functions = $1 + 1 = 2$ (for $n \ge 1$).

The number of surjective functions is the total number of functions minus the number of non-surjective functions.

Number of surjections = (Total number of functions) - (Number of non-surjective functions)

Number of surjections = $2^n - 2$.

This formula is valid for $n \ge 1$. If $n=0$, A is empty, the only function is the empty function with range $\emptyset$. This is not equal to B={a,b}, so there are 0 surjections. $2^0-2 = 1-2 = -1$, so the formula requires $n \ge 1$. Given $A = \{1, 2, ..., n\}$, it is implied that $n$ is a positive integer.

The number of surjections from A into B is $2^n - 2$.

The correct option is (B).

Solution:

The given function is $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$.

Checking if the function is defined:

The function $f(x) = \frac{1}{x}$ is defined only when the denominator is not equal to zero. That is, $x \neq 0$.

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

However, the function $f(x) = \frac{1}{x}$ is not defined for $x = 0$, and $0 \in \mathbb{R}$.

Since the function is not defined for all elements in the domain $\mathbb{R}$, the function $f: \mathbb{R} \to \mathbb{R}$ defined as $f(x) = \frac{1}{x}$ is not a valid function from $\mathbb{R}$ to $\mathbb{R}$.

Thus, the function $f$ is not defined on the entire domain $\mathbb{R}$.

Based on this, the function as stated is not defined.

The correct option is (D).

Note: If the function was defined as $f: \mathbb{R} \setminus \{0\} \to \mathbb{R}$, then it would be one-one because if $f(x_1) = f(x_2)$, then $\frac{1}{x_1} = \frac{1}{x_2}$, which implies $x_1 = x_2$ (for $x_1, x_2 \neq 0$). However, it would still not be onto $\mathbb{R}$ because there is no $x \in \mathbb{R} \setminus \{0\}$ such that $f(x) = 0$ (since $\frac{1}{x}$ is never 0). Also, if the codomain was $\mathbb{R} \setminus \{0\}$, then it would be a bijection $f: \mathbb{R} \setminus \{0\} \to \mathbb{R} \setminus \{0\}$. But the question specifies the codomain as $\mathbb{R}$.

Solution:

We are given two functions $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3x^2 - 5$ and $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = \frac{x}{x^2 + 1}$.

We need to find the composite function $g \circ f$, which is defined as $(g \circ f)(x) = g(f(x))$.

Substitute $f(x)$ into the expression for $g(x)$. The expression for $g(x)$ is $\frac{x}{x^2 + 1}$.

Replace every instance of $x$ in $g(x)$ with $f(x) = 3x^2 - 5$.

So, $(g \circ f)(x) = g(f(x)) = g(3x^2 - 5)$.

Using the definition of $g$, we have:

$(g \circ f)(x) = \frac{3x^2 - 5}{(3x^2 - 5)^2 + 1}$

Now, we need to simplify the denominator $(3x^2 - 5)^2 + 1$.

Expand the square term $(3x^2 - 5)^2$ using the formula $(a-b)^2 = a^2 - 2ab + b^2$ where $a = 3x^2$ and $b = 5$.

$(3x^2 - 5)^2 = (3x^2)^2 - 2(3x^2)(5) + (5)^2$

$(3x^2 - 5)^2 = 9x^4 - 30x^2 + 25$

Now, add 1 to this expression:

$(3x^2 - 5)^2 + 1 = (9x^4 - 30x^2 + 25) + 1$

$(3x^2 - 5)^2 + 1 = 9x^4 - 30x^2 + 26$

Substitute this back into the expression for $(g \circ f)(x)$:

$(g \circ f)(x) = \frac{3x^2 - 5}{9x^4 - 30x^2 + 26}$

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

The correct option is (A).

Solution:

We are given four functions from the set of integers $\mathbb{Z}$ to the set of integers $\mathbb{Z}$. We need to determine which of these functions is a bijection.

A function is a bijection if it is both one-one (injective) and onto (surjective).

Let's examine each function:

(A) $f(x) = x^3$

Domain: $\mathbb{Z}$, Codomain: $\mathbb{Z}$

Check for one-one:

Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{Z}$.

$x_1^3 = x_2^3$

Taking the cube root of both sides, we get $x_1 = x_2$.

So, $f(x) = x^3$ is one-one.

Check for onto:

For the function to be onto, for every $y \in \mathbb{Z}$ in the codomain, there must exist an $x \in \mathbb{Z}$ in the domain such that $f(x) = y$, i.e., $x^3 = y$.

Consider $y = 2 \in \mathbb{Z}$. Is there an integer $x$ such that $x^3 = 2$? No, since $\sqrt[3]{2}$ is not an integer.

Consider $y = 4 \in \mathbb{Z}$. Is there an integer $x$ such that $x^3 = 4$? No, since $\sqrt[3]{4}$ is not an integer.

The range of $f(x) = x^3$ for $x \in \mathbb{Z}$ is $\{..., -8, -1, 0, 1, 8, ...\}$, which is not equal to the entire codomain $\mathbb{Z}$.

So, $f(x) = x^3$ is not onto.

Therefore, $f(x) = x^3$ is not a bijection.

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

Domain: $\mathbb{Z}$, Codomain: $\mathbb{Z}$

Check for one-one:

Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{Z}$.

$x_1 + 2 = x_2 + 2$

Subtracting 2 from both sides, we get $x_1 = x_2$.

So, $f(x) = x + 2$ is one-one.

Check for onto:

For the function to be onto, for every $y \in \mathbb{Z}$ in the codomain, there must exist an $x \in \mathbb{Z}$ in the domain such that $f(x) = y$, i.e., $x + 2 = y$.

Solving for $x$, we get $x = y - 2$.

For any integer $y$, $y - 2$ is also an integer. So, for every $y \in \mathbb{Z}$, there exists an $x = y - 2 \in \mathbb{Z}$ such that $f(x) = f(y - 2) = (y - 2) + 2 = y$.

So, $f(x) = x + 2$ is onto.

Since $f(x) = x + 2$ is both one-one and onto, it is a bijection.

(C) $f(x) = 2x + 1$

Domain: $\mathbb{Z}$, Codomain: $\mathbb{Z}$

Check for one-one:

Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{Z}$.

$2x_1 + 1 = 2x_2 + 1$

Subtracting 1 from both sides, we get $2x_1 = 2x_2$.

Dividing by 2, we get $x_1 = x_2$.

So, $f(x) = 2x + 1$ is one-one.

Check for onto:

For the function to be onto, for every $y \in \mathbb{Z}$ in the codomain, there must exist an $x \in \mathbb{Z}$ in the domain such that $f(x) = y$, i.e., $2x + 1 = y$.

Solving for $x$, we get $2x = y - 1$, so $x = \frac{y - 1}{2}$.

For $x$ to be an integer, $y - 1$ must be an even number. This means $y$ must be an odd integer.

If we choose an even integer for $y$, say $y = 2 \in \mathbb{Z}$, then $x = \frac{2 - 1}{2} = \frac{1}{2}$, which is not an integer. So, there is no integer $x$ such that $f(x) = 2$.

The range of $f(x) = 2x + 1$ for $x \in \mathbb{Z}$ is the set of all odd integers, which is a proper subset of $\mathbb{Z}$.

So, $f(x) = 2x + 1$ is not onto.

Therefore, $f(x) = 2x + 1$ is not a bijection.

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

Domain: $\mathbb{Z}$, Codomain: $\mathbb{Z}$

Check for one-one:

Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{Z}$.

$x_1^2 + 1 = x_2^2 + 1$

$x_1^2 = x_2^2$

Taking the square root of both sides gives $|x_1| = |x_2|$. This implies $x_1 = x_2$ or $x_1 = -x_2$.

For example, let $x_1 = 1$ and $x_2 = -1$. Both are integers. $f(1) = 1^2 + 1 = 2$ and $f(-1) = (-1)^2 + 1 = 2$.

We have $f(1) = f(-1)$ but $1 \neq -1$.

So, $f(x) = x^2 + 1$ is not one-one.

Since the function is not one-one, it cannot be a bijection.

We don't need to check for onto, but for completeness:

Check for onto:

For the function to be onto, for every $y \in \mathbb{Z}$ in the codomain, there must exist an $x \in \mathbb{Z}$ in the domain such that $f(x) = y$, i.e., $x^2 + 1 = y$.

Solving for $x^2$, we get $x^2 = y - 1$. For $x$ to be an integer, $y-1$ must be a non-negative perfect square (0, 1, 4, 9, ...).

Consider $y = 3 \in \mathbb{Z}$. $x^2 = 3 - 1 = 2$. There is no integer $x$ such that $x^2 = 2$.

Consider $y = -2 \in \mathbb{Z}$. $x^2 = -2 - 1 = -3$. There is no real $x$ such that $x^2 = -3$, so there is no integer $x$.

The range of $f(x) = x^2 + 1$ for $x \in \mathbb{Z}$ is the set $\{1, 2, 5, 10, 17, 26, ...\}$, which is a proper subset of $\mathbb{Z}$.

So, $f(x) = x^2 + 1$ is not onto.

Therefore, $f(x) = x^2 + 1$ is not a bijection.

Based on the analysis, only the function $f(x) = x + 2$ is a bijection from $\mathbb{Z}$ to $\mathbb{Z}$.

The correct option is (B).

Solution:

Given:

The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3 + 5$.

To Find:

The inverse function $f^{-1}(x)$.

To find the inverse function, we follow these steps:

1. Replace $f(x)$ with $y$.

2. Swap $x$ and $y$.

3. Solve the new equation for $y$ in terms of $x$.

4. Replace $y$ with $f^{-1}(x)$.

Let $y = f(x)$.

$y = x^3 + 5$

Swap $x$ and $y$:

$x = y^3 + 5$

Solve for $y$:

Subtract 5 from both sides:

$y^3 = x - 5$

Take the cube root of both sides:

$y = \sqrt[3]{x - 5}$

This can also be written using fractional exponents:

$y = (x - 5)^{\frac{1}{3}}$

Replace $y$ with $f^{-1}(x)$:

$f^{-1}(x) = (x - 5)^{\frac{1}{3}}$

The inverse function of $f(x) = x^3 + 5$ is $f^{-1}(x) = (x - 5)^{\frac{1}{3}}$.

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

The correct option is (B).

Solution:

Given:

Two functions $f: A \to B$ and $g: B \to C$.

Both functions $f$ and $g$ are bijective (one-one and onto).

To Find:

The inverse of the composite function $(g \circ f)^{-1}$.

Since $f: A \to B$ and $g: B \to C$ are bijective functions, their inverses exist and are also bijective functions.

The inverse of $f$ is $f^{-1}: B \to A$.

The inverse of $g$ is $g^{-1}: C \to B$.

The composite function $g \circ f$ is defined as $(g \circ f)(x) = g(f(x))$. This function maps from set A to set C, i.e., $(g \circ f): A \to C$.

Since the composition of two bijections is also a bijection, $(g \circ f)$ is a bijective function from A to C. Therefore, its inverse $(g \circ f)^{-1}$ exists and is a function from C to A.

We need to find the expression for $(g \circ f)^{-1}$.

Let $y \in C$ be the image of some $x \in A$ under the composition $(g \circ f)$.

So, $y = (g \circ f)(x) = g(f(x))$.

Let $u = f(x)$. Then $y = g(u)$.

To find the inverse of $(g \circ f)$, we apply the inverse functions in reverse order.

From $y = g(u)$, since $g$ is bijective, we can apply $g^{-1}$ to both sides:

$g^{-1}(y) = g^{-1}(g(u))$

$g^{-1}(y) = u$

Now, substitute back $u = f(x)$:

$g^{-1}(y) = f(x)$

From $g^{-1}(y) = f(x)$, since $f$ is bijective, we can apply $f^{-1}$ to both sides:

$f^{-1}(g^{-1}(y)) = f^{-1}(f(x))$

$f^{-1}(g^{-1}(y)) = x$

Since $y = (g \circ f)(x)$, the inverse function $(g \circ f)^{-1}$ maps $y$ back to $x$.

So, $(g \circ f)^{-1}(y) = x$.

Comparing this with the previous result $f^{-1}(g^{-1}(y)) = x$, we have:

$(g \circ f)^{-1}(y) = f^{-1}(g^{-1}(y))$

Since this holds for any $y \in C$, we can write the relation between the functions as:

$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$

This is a standard property of inverse functions of composite functions.

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

The correct option is (C).

Solution:

The given function is $f : \mathbb{R} - \left\{ \frac{3}{5} \right\} \to \mathbb{R}$ defined by $f (x) = \frac{3x + 2}{5x − 3}$.

We need to find the inverse function $f^{-1}(x)$. To do this, we set $y = f(x)$, swap $x$ and $y$, and solve for $y$.

Let $y = \frac{3x + 2}{5x − 3}$.

Swapping $x$ and $y$, we get:

$x = \frac{3y + 2}{5y − 3}$

Now, we solve for $y$ in terms of $x$. Multiply both sides by $(5y - 3)$:

$x(5y - 3) = 3y + 2$

$5xy - 3x = 3y + 2$

Group terms containing $y$ on one side and terms without $y$ on the other side:

$5xy - 3y = 3x + 2$

$y(5x - 3) = 3x + 2$

Divide by $(5x - 3)$ to solve for $y$. Note that the inverse function will be defined for $x \in \mathbb{R}$ such that $5x - 3 \neq 0$, i.e., $x \neq \frac{3}{5}$. This corresponds to the range of the original function $f$.

$y = \frac{3x + 2}{5x − 3}$

Replacing $y$ with $f^{-1}(x)$, we get:

$f^{-1}(x) = \frac{3x + 2}{5x − 3}$

Comparing the expression for $f^{-1}(x)$ with the expression for $f(x)$, we observe that:

$f^{-1}(x) = \frac{3x + 2}{5x − 3}$

$f(x) = \frac{3x + 2}{5x − 3}$

Thus, $f^{-1}(x) = f(x)$. This matches option (A).

Let's also check the composition $(f \circ f)(x) = f(f(x))$.

$(f \circ f)(x) = f\left(\frac{3x + 2}{5x − 3}\right)$

Substitute $f(x)$ into the expression for $f(x)$:

$(f \circ f)(x) = \frac{3\left(\frac{3x + 2}{5x − 3}\right) + 2}{5\left(\frac{3x + 2}{5x − 3}\right) − 3}$

To simplify, multiply the numerator and the denominator by $(5x - 3)$:

$(f \circ f)(x) = \frac{3(3x + 2) + 2(5x − 3)}{5(3x + 2) − 3(5x − 3)}$

$(f \circ f)(x) = \frac{9x + 6 + 10x − 6}{15x + 10 − 15x + 9}$

$(f \circ f)(x) = \frac{(9x + 10x) + (6 - 6)}{(15x - 15x) + (10 + 9)}$

$(f \circ f)(x) = \frac{19x}{19}$

$(f \circ f)(x) = x$

Since $(f \circ f)(x) = x$, the function $f$ is its own inverse, which means $f^{-1}(x) = f(x)$. This confirms our result from finding the inverse directly and aligns with option (A).

Option (C) states $(f \circ f)(x) = -x$, which is incorrect based on our calculation $(f \circ f)(x) = x$.

Options (B) and (D) are also incorrect as $f^{-1}(x) = f(x)$.

The correct option is (A).

Solution:

The given function is $f : [0, 1] \to [0, 1]$ defined as:

$f (x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1−x, & \text{if } x \text{ is irrational} \end{cases}$

We need to find the composite function $(f \circ f)(x)$, which is defined as $(f \circ f)(x) = f(f(x))$.

We consider two cases based on whether $x \in [0, 1]$ is rational or irrational.

Case 1: $x$ is a rational number in $[0, 1]$.

If $x$ is rational, by the definition of $f$, we have $f(x) = x$.

Now we evaluate $f(f(x))$, which is $f(x)$.

Since $x$ is rational, $f(x) = x$.

So, if $x$ is rational, $(f \circ f)(x) = x$.

Case 2: $x$ is an irrational number in $[0, 1]$.

If $x$ is irrational, by the definition of $f$, we have $f(x) = 1 - x$.

Now we need to evaluate $f(f(x))$, which is $f(1 - x)$.

We need to determine if $1 - x$ is rational or irrational. Let's assume, for the sake of contradiction, that $1 - x$ is rational.

If $1 - x$ is rational, then $1 - x = q$ for some rational number $q$.

Rearranging the equation, we get $x = 1 - q$.

Since 1 is a rational number and $q$ is a rational number, their difference $1 - q$ is also a rational number.

This means $x$ is rational, which contradicts our initial assumption that $x$ is irrational.

Therefore, if $x$ is an irrational number, then $1 - x$ must also be an irrational number.

Since $1 - x$ is irrational (when $x$ is irrational), we use the second part of the definition of $f$ to evaluate $f(1 - x)$:

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

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

$f(1 - x) = x$

So, if $x$ is irrational, $(f \circ f)(x) = x$.

Combining both cases, we see that for any $x \in [0, 1]$, whether rational or irrational, we have $(f \circ f)(x) = x$.

Thus, $(f \circ f)(x) = x$ for all $x \in [0, 1]$.

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

The correct option is (C).

Solution:

Given:

The function $f : [2, \infty) \to \mathbb{R}$ defined by $f(x) = x^2 - 4x + 5$.

To Find:

The range of the function $f$.

The function is a quadratic function $f(x) = x^2 - 4x + 5$. The graph of this function is a parabola.

To find the range, we can find the vertex of the parabola.

For a quadratic function $ax^2 + bx + c$, the x-coordinate of the vertex is given by $x = -\frac{b}{2a}$.

In this case, $a=1$, $b=-4$, $c=5$.

The x-coordinate of the vertex is $x = -\frac{(-4)}{2(1)} = \frac{4}{2} = 2$.

The y-coordinate (function value) at the vertex is $f(2)$.

$f(2) = (2)^2 - 4(2) + 5$

$f(2) = 4 - 8 + 5$

$f(2) = 1$

The vertex of the parabola is at the point $(2, 1)$.

The domain of the function is given as $[2, \infty)$. This interval starts exactly at the x-coordinate of the vertex.

Since the coefficient of $x^2$ is $1 > 0$, the parabola opens upwards.

This means that the function has a minimum value at the vertex.

Considering the domain $[2, \infty)$, the smallest value of $x$ is 2, which is the x-coordinate of the vertex. The function value at this point is $f(2) = 1$.

As $x$ increases from 2 towards $\infty$, the function values $f(x)$ will increase because the parabola opens upwards and we are considering the part of the domain to the right of the vertex.

As $x \to \infty$, $f(x) = x^2 - 4x + 5 \to \infty$.

Therefore, the range of the function $f$ on the domain $[2, \infty)$ starts from the minimum value at the vertex (inclusive) and extends to $\infty$.

The minimum value is 1.

The range is $[1, \infty)$.

Alternatively, we can complete the square for $f(x)$:

$f(x) = x^2 - 4x + 5$

$f(x) = (x^2 - 4x + 4) + 1$

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

Since the domain is $[2, \infty)$, we have $x \ge 2$.

Subtracting 2 from all parts of the inequality, we get $x - 2 \ge 0$.

Squaring both sides (since $x-2 \ge 0$), we get $(x - 2)^2 \ge 0^2$, so $(x - 2)^2 \ge 0$.

Adding 1 to both sides, we get $(x - 2)^2 + 1 \ge 0 + 1$, so $(x - 2)^2 + 1 \ge 1$.

Since $f(x) = (x - 2)^2 + 1$, this means $f(x) \ge 1$.

The smallest value that $f(x)$ can take is 1, which occurs when $x - 2 = 0$, i.e., $x = 2$. Since $x=2$ is in the domain, the minimum value is 1.

As $x$ increases in the domain $[2, \infty)$, $(x-2)^2$ increases, and thus $f(x)$ increases without bound.

The range of $f$ is $[1, \infty)$.

The correct option is (B).

Solution:

Given:

Function $f : \mathbb{N} \to \mathbb{R}$ defined by $f(x) = \frac{2x − 1}{2}$.

Function $g : \mathbb{Q} \to \mathbb{R}$ defined by $g(x) = x + 2$.

To Find:

The value of $(g \circ f)\left(\frac{3}{2}\right)$.

The composite function $(g \circ f)$ is defined as $(g \circ f)(x) = g(f(x))$.

The domain of the composite function $(g \circ f)$ is the set of all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$.

The domain of $f$ is $\mathbb{N}$, the set of natural numbers (which typically starts with 1, i.e., $\{1, 2, 3, ...\}$ or $\{0, 1, 2, 3, ...\}$ depending on convention, but in either case consists of non-fractional, non-negative integers).

The domain of $g$ is $\mathbb{Q}$, the set of rational numbers.

For any $x \in \mathbb{N}$, the value $f(x) = \frac{2x - 1}{2} = x - \frac{1}{2}$.

Since $x$ is an integer, $x - \frac{1}{2}$ is always a rational number. Thus, $f(x) \in \mathbb{Q}$ for all $x \in \mathbb{N}$.

Therefore, the domain of the composite function $(g \circ f)$ is $\mathbb{N}$.

We are asked to evaluate $(g \circ f)\left(\frac{3}{2}\right)$. The input value is $\frac{3}{2}$.

We need to check if this input value $\frac{3}{2}$ is in the domain of $(g \circ f)$, which is $\mathbb{N}$.

The number $\frac{3}{2}$ is equal to 1.5. This is a rational number ($\frac{3}{2} \in \mathbb{Q}$), but it is not a natural number ($\frac{3}{2} \notin \mathbb{N}$).

Since the input value $x = \frac{3}{2}$ is not in the domain of the composite function $(g \circ f)$, the expression $(g \circ f)\left(\frac{3}{2}\right)$ is not defined.

Based on the definition of the functions and their domains as given in the question, the composite function $(g \circ f)$ is only defined for natural numbers. Evaluating it at $\frac{3}{2}$ is therefore not possible within the specified domain.

The correct option is (D) none of these, as the expression is not defined.

Solution:

The given piecewise function is defined as:

$f(x) = \begin{cases} 2x & : x > 3 \\ x^2 & : 1 < x \leq 3 \\ 3x & : x \leq 1 \end{cases}$

We need to evaluate $f(-1)$, $f(2)$, and $f(4)$ using the appropriate definition of the function for each input value.

Evaluate $f(-1)$:

We check which interval $-1$ belongs to:

$-1 > 3$ is False.

$1 < -1 \leq 3$ is False.

$-1 \leq 1$ is True.

Since $-1 \leq 1$, we use the third part of the definition: $f(x) = 3x$.

$f(-1) = 3 \times (-1) = -3$.

Evaluate $f(2)$:

We check which interval $2$ belongs to:

$2 > 3$ is False.

$1 < 2 \leq 3$ is True.

Since $1 < 2 \leq 3$, we use the second part of the definition: $f(x) = x^2$.

$f(2) = (2)^2 = 4$.

Evaluate $f(4)$:

We check which interval $4$ belongs to:

$4 > 3$ is True.

Since $4 > 3$, we use the first part of the definition: $f(x) = 2x$.

$f(4) = 2 \times 4 = 8$.

Now we need to find the sum $f(-1) + f(2) + f(4)$.

Sum $= f(-1) + f(2) + f(4) = (-3) + 4 + 8$.

Sum $= 1 + 8 = 9$.

The value of $f(-1) + f(2) + f(4)$ is 9.

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

The correct option is (A).

Solution:

Given:

The function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \tan x$.

To Find:

The value of $f^{-1}(1)$.

For a function to have an inverse function, it must be bijective (both one-one and onto) from its domain to its codomain.

Let's analyze the given function $f(x) = \tan x$ with domain $\mathbb{R}$ and codomain $\mathbb{R}$.

The function $\tan x = \frac{\sin x}{\cos x}$ is not defined for all real numbers $x$. It is undefined when $\cos x = 0$, which occurs at $x = \frac{\pi}{2} + n\pi$ for any integer $n \in \mathbb{Z}$. Since $\mathbb{R}$ includes these values, the function $f(x) = \tan x$ as stated, with domain $\mathbb{R}$, is not even well-defined on its entire domain.

Even if we were to consider the natural domain of $\tan x$ (i.e., $\mathbb{R} \setminus \left\{ \frac{\pi}{2} + n\pi : n \in \mathbb{Z} \right\}$), the function $f(x) = \tan x$ is not one-one on this domain because it is periodic with period $\pi$. For example, $\tan\left(\frac{\pi}{4}\right) = 1$ and $\tan\left(\frac{5\pi}{4}\right) = 1$, but $\frac{\pi}{4} \neq \frac{5\pi}{4}$.

Since the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \tan x$ is neither well-defined on its entire domain nor one-one, it is not bijective. Therefore, the inverse function $f^{-1} : \mathbb{R} \to \mathbb{R}$ does not exist.

If the inverse function $f^{-1}$ does not exist, then $f^{-1}(1)$ (as an output of the inverse function) does not exist.

Note that the set of values $x$ for which $\tan x = 1$ is $\{x \in \mathbb{R} \mid \tan x = 1\}$. The general solution for $\tan x = 1$ is $x = \frac{\pi}{4} + n\pi$, where $n \in \mathbb{Z}$. This set is $\left\{ \frac{\pi}{4}, \frac{5\pi}{4}, -\frac{3\pi}{4}, ... \right\}$. Option (B) represents this set of preimages of 1. However, $f^{-1}(1)$ usually denotes a single value output by the inverse *function*. Since the inverse function does not exist, this interpretation is not valid here.

Based on the fact that the function $f: \mathbb{R} \to \mathbb{R}, f(x) = \tan x$ is not invertible, the value $f^{-1}(1)$ does not exist in the sense of the output of an inverse function from $\mathbb{R}$ to $\mathbb{R}$.

The correct option is (C).

Given:

Relation R is defined in $\mathbb{N}$ by aRb if $2a + 3b = 30$.

Solution:

The relation R contains ordered pairs $(a, b)$ such that $a$ and $b$ are natural numbers ($\mathbb{N}$) and they satisfy the equation $2a + 3b = 30$. We assume $\mathbb{N} = \{1, 2, 3, ...\}$.

We need to find pairs $(a, b)$ of natural numbers such that $2a + 3b = 30$.

From the equation, $2a = 30 - 3b$.

Since $a \in \mathbb{N}$, $a \geq 1$. Therefore, $2a \geq 2$.

Substituting this into the equation:

So, $b \leq 9.\overline{3}$.

Also, from $2a = 30 - 3b$, since $2a$ is an even number and 30 is an even number, $3b$ must be an even number. This implies that $b$ must be an even number.

Since $b$ is a natural number, $b \geq 1$. Combining this with $b \leq 9.\overline{3}$ and the requirement that $b$ is even, the possible values for $b$ are 2, 4, 6, 8.

Let's find the corresponding values of $a$ for each of these $b$ values using the equation $a = \frac{30 - 3b}{2}$:

When $b=2$:

Pair: $(12, 2)$. Since $12 \in \mathbb{N}$, this pair is in R.

When $b=4$:

Pair: $(9, 4)$. Since $9 \in \mathbb{N}$, this pair is in R.

When $b=6$:

Pair: $(6, 6)$. Since $6 \in \mathbb{N}$, this pair is in R.

When $b=8$:

Pair: $(3, 8)$. Since $3 \in \mathbb{N}$, this pair is in R.

If we take the next even natural number for $b$, which is 10:

When $b=10$:

Pair: $(0, 10)$. Since $0 \notin \mathbb{N}$ (assuming $\mathbb{N} = \{1, 2, 3, ...\}$), this pair is not in R.

The relation R is the set of all such ordered pairs $(a, b)$.

So, R = $\{(12, 2), (9, 4), (6, 6), (3, 8)\}$.

Given:

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

Relation $R$ is defined on $A$ by $R = \{(a, b) : |a^2 - b^2| < 8$, where $a, b \in A\}$.

Solution:

We need to find all ordered pairs $(a, b)$ such that $a$ and $b$ are elements of set $A = \{1, 2, 3, 4, 5\}$, and the absolute difference between the square of $a$ and the square of $b$ is less than 8.

First, let's list the squares of the elements in set $A$:

$1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

Now, we examine each possible pair $(a, b)$ with $a \in A$ and $b \in A$ and check the condition $|a^2 - b^2| < 8$.

For $a=1$:

• $(1, 1)$: $|1^2 - 1^2| = |1 - 1| = 0$. $0 < 8$. Include $(1, 1)$.
• $(1, 2)$: $|1^2 - 2^2| = |1 - 4| = |-3| = 3$. $3 < 8$. Include $(1, 2)$.
• $(1, 3)$: $|1^2 - 3^2| = |1 - 9| = |-8| = 8$. $8 \not< 8$. Exclude $(1, 3)$.
• $(1, 4)$: $|1^2 - 4^2| = |1 - 16| = |-15| = 15$. $15 \not< 8$. Exclude $(1, 4)$.
• $(1, 5)$: $|1^2 - 5^2| = |1 - 25| = |-24| = 24$. $24 \not< 8$. Exclude $(1, 5)$.

For $a=2$:

• $(2, 1)$: $|2^2 - 1^2| = |4 - 1| = 3$. $3 < 8$. Include $(2, 1)$.
• $(2, 2)$: $|2^2 - 2^2| = |4 - 4| = 0$. $0 < 8$. Include $(2, 2)$.
• $(2, 3)$: $|2^2 - 3^2| = |4 - 9| = |-5| = 5$. $5 < 8$. Include $(2, 3)$.
• $(2, 4)$: $|2^2 - 4^2| = |4 - 16| = |-12| = 12$. $12 \not< 8$. Exclude $(2, 4)$.
• $(2, 5)$: $|2^2 - 5^2| = |4 - 25| = |-21| = 21$. $21 \not< 8$. Exclude $(2, 5)$.

For $a=3$:

• $(3, 1)$: $|3^2 - 1^2| = |9 - 1| = 8$. $8 \not< 8$. Exclude $(3, 1)$.
• $(3, 2)$: $|3^2 - 2^2| = |9 - 4| = 5$. $5 < 8$. Include $(3, 2)$.
• $(3, 3)$: $|3^2 - 3^2| = |9 - 9| = 0$. $0 < 8$. Include $(3, 3)$.
• $(3, 4)$: $|3^2 - 4^2| = |9 - 16| = |-7| = 7$. $7 < 8$. Include $(3, 4)$.
• $(3, 5)$: $|3^2 - 5^2| = |9 - 25| = |-16| = 16$. $16 \not< 8$. Exclude $(3, 5)$.

For $a=4$:

• $(4, 1)$: $|4^2 - 1^2| = |16 - 1| = 15$. $15 \not< 8$. Exclude $(4, 1)$.
• $(4, 2)$: $|4^2 - 2^2| = |16 - 4| = 12$. $12 \not< 8$. Exclude $(4, 2)$.
• $(4, 3)$: $|4^2 - 3^2| = |16 - 9| = 7$. $7 < 8$. Include $(4, 3)$.
• $(4, 4)$: $|4^2 - 4^2| = |16 - 16| = 0$. $0 < 8$. Include $(4, 4)$.
• $(4, 5)$: $|4^2 - 5^2| = |16 - 25| = |-9| = 9$. $9 \not< 8$. Exclude $(4, 5)$.

For $a=5$:

• $(5, 1)$: $|5^2 - 1^2| = |25 - 1| = 24$. $24 \not< 8$. Exclude $(5, 1)$.
• $(5, 2)$: $|5^2 - 2^2| = |25 - 4| = 21$. $21 \not< 8$. Exclude $(5, 2)$.
• $(5, 3)$: $|5^2 - 3^2| = |25 - 9| = 16$. $16 \not< 8$. Exclude $(5, 3)$.
• $(5, 4)$: $|5^2 - 4^2| = |25 - 16| = 9$. $9 \not< 8$. Exclude $(5, 4)$.
• $(5, 5)$: $|5^2 - 5^2| = |25 - 25| = 0$. $0 < 8$. Include $(5, 5)$.

The pairs $(a, b)$ that satisfy the condition $|a^2 - b^2| < 8$ are the ones we included.

Therefore, the relation $R$ is given by:

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

Given:

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

Function $g = \{(2, 3), (5, 1), (1, 3)\}$

To Find:

$g \circ f$ and $f \circ g$

Solution:

The composite function $g \circ f$ is defined as $(g \circ f)(x) = g(f(x))$ for all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$.

The domain of $f$ is $\{1, 3, 4\}$. The range of $f$ is $\{1, 2, 5\}$.

The domain of $g$ is $\{1, 2, 5\}$. The range of $g$ is $\{1, 3\}$.

For $g \circ f$ to be defined, the range of $f$ must be a subset of the domain of $g$. Here, the range of $f$ is $\{1, 2, 5\}$ and the domain of $g$ is $\{1, 2, 5\}$. Since $\{1, 2, 5\} \subseteq \{1, 2, 5\}$, $g \circ f$ is defined for all elements in the domain of $f$.

Let's find $g(f(x))$ for each $x$ in the domain of $f$:

• For $x=1$: $f(1) = 2$. Then $g(f(1)) = g(2) = 3$. So, $(1, 3) \in g \circ f$.
• For $x=3$: $f(3) = 5$. Then $g(f(3)) = g(5) = 1$. So, $(3, 1) \in g \circ f$.
• For $x=4$: $f(4) = 1$. Then $g(f(4)) = g(1) = 3$. So, $(4, 3) \in g \circ f$.

Thus, $g \circ f = \{(1, 3), (3, 1), (4, 3)\}$.

The composite function $f \circ g$ is defined as $(f \circ g)(x) = f(g(x))$ for all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.

The domain of $g$ is $\{1, 2, 5\}$. The range of $g$ is $\{1, 3\}$.

The domain of $f$ is $\{1, 3, 4\}$. The range of $f$ is $\{1, 2, 5\}$.

For $f \circ g$ to be defined, the range of $g$ must be a subset of the domain of $f$. Here, the range of $g$ is $\{1, 3\}$ and the domain of $f$ is $\{1, 3, 4\}$. Since $\{1, 3\} \subseteq \{1, 3, 4\}$, $f \circ g$ is defined for all elements in the domain of $g$.

Let's find $f(g(x))$ for each $x$ in the domain of $g$:

• For $x=1$: $g(1) = 3$. Then $f(g(1)) = f(3) = 5$. So, $(1, 5) \in f \circ g$.
• For $x=2$: $g(2) = 3$. Then $f(g(2)) = f(3) = 5$. So, $(2, 5) \in f \circ g$.
• For $x=5$: $g(5) = 1$. Then $f(g(5)) = f(1) = 2$. So, $(5, 2) \in f \circ g$.

Thus, $f \circ g = \{(1, 5), (2, 5), (5, 2)\}$.

Answer:

$g \circ f = \{(1, 3), (3, 1), (4, 3)\}$

$f \circ g = \{(1, 5), (2, 5), (5, 2)\}$

Given:

$f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{x}{\sqrt{1 + x^2}}$.

To Find:

$(f \circ f \circ f)(x)$

Solution:

First, we find $(f \circ f)(x) = f(f(x))$:

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

Substitute $\frac{x}{\sqrt{1 + x^2}}$ for $x$ in the expression for $f(x)$:

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

Simplify the term inside the square root in the denominator:

$\left(\frac{x}{\sqrt{1 + x^2}}\right)^2 = \frac{x^2}{1 + x^2}$

The denominator is $\sqrt{1 + \frac{x^2}{1 + x^2}} = \sqrt{\frac{1 + x^2 + x^2}{1 + x^2}} = \sqrt{\frac{1 + 2x^2}{1 + x^2}} = \frac{\sqrt{1 + 2x^2}}{\sqrt{1 + x^2}}$

So, $(f \circ f)(x) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\frac{\sqrt{1 + 2x^2}}{\sqrt{1 + x^2}}} = \frac{x}{\sqrt{1 + x^2}} \times \frac{\sqrt{1 + x^2}}{\sqrt{1 + 2x^2}} = \frac{x}{\sqrt{1 + 2x^2}}$

Thus, $(f \circ f)(x) = \frac{x}{\sqrt{1 + 2x^2}}$.

Now, we find $(f \circ f \circ f)(x) = f((f \circ f)(x))$:

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

Substitute $\frac{x}{\sqrt{1 + 2x^2}}$ for $x$ in the expression for $f(x)$:

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

Simplify the term inside the square root in the denominator:

$\left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2 = \frac{x^2}{1 + 2x^2}$

The denominator is $\sqrt{1 + \frac{x^2}{1 + 2x^2}} = \sqrt{\frac{1 + 2x^2 + x^2}{1 + 2x^2}} = \sqrt{\frac{1 + 3x^2}{1 + 2x^2}} = \frac{\sqrt{1 + 3x^2}}{\sqrt{1 + 2x^2}}$

So, $(f \circ f \circ f)(x) = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\frac{\sqrt{1 + 3x^2}}{\sqrt{1 + 2x^2}}} = \frac{x}{\sqrt{1 + 2x^2}} \times \frac{\sqrt{1 + 2x^2}}{\sqrt{1 + 3x^2}} = \frac{x}{\sqrt{1 + 3x^2}}$

Answer:

$(f \circ f \circ f)(x) = \frac{x}{\sqrt{1 + 3x^2}}$

Given:

Function $f(x) = 4 - (x - 7)^3$.

To Find:

$f^{-1}(x)$

Solution:

To find the inverse function, we set $y = f(x)$ and solve for $x$ in terms of $y$. Then we swap $x$ and $y$ to get $f^{-1}(x)$.

Let $y = f(x)$.

$y = 4 - (x - 7)^3$

Now, we solve for $x$:

Subtract 4 from both sides:

$y - 4 = - (x - 7)^3$

Multiply both sides by -1:

$-(y - 4) = (x - 7)^3$

$4 - y = (x - 7)^3$

Take the cube root of both sides:

$\sqrt[3]{4 - y} = x - 7$

Add 7 to both sides:

$x = 7 + \sqrt[3]{4 - y}$

Now, swap $x$ and $y$ to find the inverse function $f^{-1}(x)$:

$f^{-1}(x) = 7 + \sqrt[3]{4 - x}$

Answer:

$f^{-1}(x) = 7 + \sqrt[3]{4 - x}$

Given:

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

Relation $R = \{(3, 1), (1, 3), (3, 3)\}$ defined on set A.

To Check:

Whether R is symmetric, transitive but not reflexive.

Analysis of Properties:

Let's check the properties of the relation R:

Reflexivity:

A relation R on set A is reflexive if $(a, a) \in R$ for every $a \in A$.

The elements in A are 1, 2, and 3.

We need to check if $(1, 1) \in R$, $(2, 2) \in R$, and $(3, 3) \in R$.

From the definition of R, we have $(3, 3) \in R$, but $(1, 1) \notin R$ and $(2, 2) \notin R$.

Since not all elements of A are related to themselves, R is not reflexive.

Symmetry:

A relation R on set A is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$ for all $a, b \in A$.

Check the pairs in R:

• $(3, 1) \in R$. Is $(1, 3) \in R$? Yes.
• $(1, 3) \in R$. Is $(3, 1) \in R$? Yes.
• $(3, 3) \in R$. Is $(3, 3) \in R$? Yes.

For every pair $(a, b)$ in R, the pair $(b, a)$ is also in R.

Thus, R is symmetric.

Transitivity:

A relation R on set A is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in A$.

Check the possible combinations:

• Consider $(3, 1) \in R$ and $(1, 3) \in R$. We must check if $(3, 3) \in R$. Yes, $(3, 3) \in R$. This case holds.
• Consider $(1, 3) \in R$ and $(3, 1) \in R$. We must check if $(1, 1) \in R$. No, $(1, 1) \notin R$. This case fails.
• Consider $(1, 3) \in R$ and $(3, 3) \in R$. We must check if $(1, 3) \in R$. Yes, $(1, 3) \in R$. This case holds.
• Consider $(3, 3) \in R$ and $(3, 1) \in R$. We must check if $(3, 1) \in R$. Yes, $(3, 1) \in R$. This case holds.

Since we found a case where $(a, b) \in R$ and $(b, c) \in R$ but $(a, c) \notin R$ (specifically, $(1, 3) \in R$ and $(3, 1) \in R$ but $(1, 1) \notin R$), the relation R is not transitive.

The statement claims that R is symmetric, transitive but not reflexive.

Our analysis shows that R is symmetric, not transitive, and not reflexive.

Since R is not transitive, the statement is false.

Answer:

False

Given:

Function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \sin(3x + 2)$ for all $x \in \mathbb{R}$.

To Check:

Whether $f$ is invertible.

Analysis:

A function is invertible if and only if it is both injective (one-to-one) and surjective (onto).

Let's examine the properties of the function $f(x) = \sin(3x + 2)$ on the domain $\mathbb{R}$ and codomain $\mathbb{R}$.

Injectivity:

A function $f$ is injective if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Consider $x_1 = 0$ and $x_2 = \frac{2\pi}{3}$. Both $x_1, x_2 \in \mathbb{R}$.

$f(x_1) = f(0) = \sin(3(0) + 2) = \sin(2)$

$f(x_2) = f\left(\frac{2\pi}{3}\right) = \sin\left(3\left(\frac{2\pi}{3}\right) + 2\right) = \sin(2\pi + 2)$

Using the property $\sin(\theta + 2\pi) = \sin(\theta)$, we have:

$f\left(\frac{2\pi}{3}\right) = \sin(2\pi + 2) = \sin(2)$

So, $f(0) = f\left(\frac{2\pi}{3}\right) = \sin(2)$, but $0 \neq \frac{2\pi}{3}$.

Since there exist different inputs that produce the same output, the function $f$ is not injective on $\mathbb{R}$.

Surjectivity:

A function $f : A \to B$ is surjective if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$.

The range of the sine function, $\sin(\theta)$, is $[-1, 1]$. Therefore, the range of $f(x) = \sin(3x + 2)$ is also $[-1, 1]$.

The codomain of $f$ is given as $\mathbb{R}$.

Since the range of $f$, which is $[-1, 1]$, is not equal to the codomain $\mathbb{R}$ (for example, there is no $x \in \mathbb{R}$ such that $f(x) = 5$), the function $f$ is not surjective.

Since the function $f(x) = \sin(3x + 2)$ is neither injective nor surjective on the given domain and codomain, it is not invertible.

The statement claims that $f$ is invertible.

Answer:

False

Statement:

Every relation which is symmetric and transitive is also reflexive.

Analysis:

Let A be a set and R be a relation on A.

A relation R is reflexive if for every element $a \in A$, the pair $(a, a)$ is in R.

A relation R is symmetric if for every pair $(a, b) \in R$, the pair $(b, a)$ is also in R.

A relation R is transitive if for every three elements $a, b, c \in A$, whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c)$ is also in R.

Consider a counterexample:

Let $A = \{1, 2\}$ and the relation R be defined as $R = \{(1, 1)\}$.

Let's check the properties of R:

Symmetry:

The only pair in R is $(1, 1)$. If we check for symmetry, we need to see if when $(a, b) \in R$, then $(b, a) \in R$. For the pair $(1, 1) \in R$, we need to check if $(1, 1) \in R$. It is. There are no other pairs to check. So, R is symmetric.

Transitivity:

We need to check if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. The only way to satisfy the premise "$(a, b) \in R$ and $(b, c) \in R$" is if $a=1, b=1, c=1$. In this case, $(1, 1) \in R$ and $(1, 1) \in R$. We need to check if $(a, c) = (1, 1)$ is in R. It is. There are no other combinations of pairs in R that satisfy the premise. So, R is transitive.

Reflexivity:

For R to be reflexive on set A = {1, 2}, it must contain $(1, 1)$ and $(2, 2)$.

The relation R contains $(1, 1)$ but it does not contain $(2, 2)$.

So, R is not reflexive on set A.

We have found a relation R on set A that is symmetric and transitive but is not reflexive on A.

Therefore, the statement "Every relation which is symmetric and transitive is also reflexive" is false.

Answer:

False

Given:

The relation R is defined on the set of integers $\mathbb{Z}$ by $mRn$ if $m$ is an integral multiple of $n$. This means $m = kn$ for some integer $k$.

To Check:

Whether the relation R is reflexive, symmetric, and transitive.

Analysis of Properties:

Let's check the properties of the relation R on $\mathbb{Z}$.

Reflexivity:

R is reflexive if for every $a \in \mathbb{Z}$, $(a, a) \in R$. This means $a$ must be an integral multiple of $a$.

Is $a = k \cdot a$ for some integer $k$, for all $a \in \mathbb{Z}$?

For any integer $a$, we can choose $k=1$. Then $a = 1 \cdot a$. This is true for all integers, including $a=0$ ($0 = 1 \cdot 0$).

Thus, $(a, a) \in R$ for all $a \in \mathbb{Z}$. The relation R is reflexive.

Symmetry:

R is symmetric if for every $(a, b) \in R$, $(b, a) \in R$. This means if $a$ is an integral multiple of $b$, then $b$ must be an integral multiple of $a$.

If $(a, b) \in R$, then $a = k_1 b$ for some integer $k_1$.

If $(b, a) \in R$, then $b = k_2 a$ for some integer $k_2$.

Consider $a=4$ and $b=2$. $4$ is an integral multiple of $2$ because $4 = 2 \times 2$. So, $(4, 2) \in R$.

For R to be symmetric, $(2, 4)$ must be in R, meaning $2$ must be an integral multiple of $4$.

Is $2 = k \cdot 4$ for some integer $k$? This would mean $k = \frac{2}{4} = \frac{1}{2}$, which is not an integer.

Therefore, $(2, 4) \notin R$.

Since there exists a pair $(4, 2) \in R$ but $(2, 4) \notin R$, the relation R is not symmetric.

Transitivity:

R is transitive if for every $(a, b) \in R$ and $(b, c) \in R$, $(a, c) \in R$. This means if $a$ is an integral multiple of $b$ and $b$ is an integral multiple of $c$, then $a$ must be an integral multiple of $c.

Assume $(a, b) \in R$ and $(b, c) \in R$.

Substitute (ii) into (i):

$a = k_1 (k_2 c) = (k_1 k_2) c$

Since $k_1$ and $k_2$ are integers, their product $k_1 k_2$ is also an integer. Let $k_3 = k_1 k_2$.

$a = k_3 c$, where $k_3$ is an integer.

This shows that $a$ is an integral multiple of $c$. Therefore, $(a, c) \in R$.

The relation R is transitive.

The statement claims that the relation R is reflexive, symmetric, and transitive. Our analysis shows that R is reflexive and transitive, but not symmetric.

Since R is not symmetric, the statement is false.

Answer:

False

Given:

Set $A = \{0, 1\}$

Set of natural numbers $\mathbb{N}$.

Function $f : \mathbb{N} \to A$ defined by:

• $f(2n - 1) = 0$, for all $n \in \mathbb{N}$
• $f(2n) = 1$, for all $n \in \mathbb{N}$

To Check:

Whether the mapping $f$ is onto (surjective).

Analysis:

A function $f: X \to Y$ is onto (surjective) if for every element $y$ in the codomain $Y$, there exists at least one element $x$ in the domain $X$ such that $f(x) = y$.

The domain of $f$ is $\mathbb{N} = \{1, 2, 3, 4, 5, ...\}$.

The codomain of $f$ is $A = \{0, 1\}$.

We need to determine if every element in the codomain $\{0, 1\}$ has a preimage in the domain $\mathbb{N}$.

Consider the elements in the codomain:

• Element 0: We need to check if there exists an $x \in \mathbb{N}$ such that $f(x) = 0$. The definition of $f$ states that $f(2n - 1) = 0$ for all $n \in \mathbb{N}$. The expression $2n - 1$ generates all odd natural numbers as $n$ takes values $1, 2, 3, ...$. Since all odd natural numbers are in the domain $\mathbb{N}$, the element 0 in the codomain has preimages in the domain (e.g., $f(1)=0$, $f(3)=0$, etc.).
• Element 1: We need to check if there exists an $x \in \mathbb{N}$ such that $f(x) = 1$. The definition of $f$ states that $f(2n) = 1$ for all $n \in \mathbb{N}$. The expression $2n$ generates all even natural numbers as $n$ takes values $1, 2, 3, ...$. Since all even natural numbers are in the domain $\mathbb{N}$, the element 1 in the codomain has preimages in the domain (e.g., $f(2)=1$, $f(4)=1$, etc.).

Since both elements in the codomain, 0 and 1, have at least one preimage in the domain $\mathbb{N}$, the function $f$ is onto.

The statement claims that the mapping $f$ is onto. Our analysis confirms that it is onto.

Answer:

True

Given:

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

Relation $R = \{(1, 1), (1, 2), (2, 1), (3, 3)\}$ defined on set A.

To Check:

Whether R is reflexive, symmetric, and transitive.

Analysis of Properties:

Let's check the properties of the relation R:

Reflexivity:

A relation R on set A is reflexive if $(a, a) \in R$ for every $a \in A$.

The elements in A are 1, 2, and 3.

We need to check if $(1, 1) \in R$, $(2, 2) \in R$, and $(3, 3) \in R$.

From the definition of R, we have $(1, 1) \in R$ and $(3, 3) \in R$. However, $(2, 2) \notin R$.

Since not all elements of A are related to themselves, R is not reflexive.

Symmetry:

A relation R on set A is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$ for all $a, b \in A$.

Check the pairs in R:

• $(1, 1) \in R$. Is $(1, 1) \in R$? Yes.
• $(1, 2) \in R$. Is $(2, 1) \in R$? Yes.
• $(2, 1) \in R$. Is $(1, 2) \in R$? Yes.
• $(3, 3) \in R$. Is $(3, 3) \in R$? Yes.

For every pair $(a, b)$ in R, the pair $(b, a)$ is also in R.

Thus, R is symmetric.

Transitivity:

A relation R on set A is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in A$.

Check the possible combinations:

• Consider $(1, 2) \in R$ and $(2, 1) \in R$. We must check if $(1, 1) \in R$. Yes, $(1, 1) \in R$. This case holds.
• Consider $(2, 1) \in R$ and $(1, 2) \in R$. We must check if $(2, 2) \in R$. No, $(2, 2) \notin R$. This case fails.

Since we found a case where $(a, b) \in R$ and $(b, c) \in R$ but $(a, c) \notin R$ (specifically, $(2, 1) \in R$ and $(1, 2) \in R$ but $(2, 2) \notin R$), the relation R is not transitive.

The statement claims that R is reflexive, symmetric, and transitive.

Our analysis shows that R is symmetric, but not reflexive and not transitive.

Since R is not reflexive and not transitive, the statement is false.

Answer:

False

Statement:

The composition of functions is commutative.

Analysis:

Function composition is an operation that takes two functions, say $f$ and $g$, and produces a new function, $f \circ g$, where $(f \circ g)(x) = f(g(x))$. Similarly, $(g \circ f)(x) = g(f(x))$.

For the composition of functions to be commutative, we would need $(f \circ g)(x) = (g \circ f)(x)$ for all $x$ in the domains of both composite functions, and the domains and codomains must align appropriately for both compositions to be defined and have the same resulting domain and codomain.

In general, function composition is not commutative. This means that for most pairs of functions $f$ and $g$, $f \circ g \neq g \circ f$.

Consider a counterexample:

Let $f(x) = x + 1$ and $g(x) = x^2$. Both functions are defined on $\mathbb{R}$.

Let's compute $(f \circ g)(x)$:

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

Now, let's compute $(g \circ f)(x)$:

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

Comparing the two results:

From (i) and (ii), we see that $x^2 + 1 \neq x^2 + 2x + 1$. For example, if $x=1$, $(f \circ g)(1) = 1^2 + 1 = 2$, while $(g \circ f)(1) = (1 + 1)^2 = 2^2 = 4$. Since $2 \neq 4$, $(f \circ g)(1) \neq (g \circ f)(1)$.

Therefore, $f \circ g \neq g \circ f$.

This counterexample demonstrates that the composition of functions is generally not commutative.

Answer:

False

Statement:

The composition of functions is associative.

Analysis:

Function composition involves combining functions sequentially. Associativity means that the grouping of functions in a composition does not affect the result, provided the compositions are defined. Specifically, for three functions $f$, $g$, and $h$, we need to check if $(f \circ g) \circ h = f \circ (g \circ h)$.

Let $x$ be an element in the domain of $h$.

Consider the left side: $((f \circ g) \circ h)(x)$.

By definition, $(f \circ g)(y) = f(g(y))$. So, $((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = f(g(h(x)))$.

Consider the right side: $(f \circ (g \circ h))(x)$.

By definition, $(g \circ h)(x) = g(h(x))$. So, $(f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(g(h(x)))$.

Since $((f \circ g) \circ h)(x) = f(g(h(x)))$ and $(f \circ (g \circ h))(x) = f(g(h(x)))$, we have $((f \circ g) \circ h)(x) = (f \circ (g \circ h))(x)$, provided all the intermediate compositions are defined.

The compositions are defined if the range of the inner function is a subset of the domain of the outer function at each step.

Let $h: A \to B$, $g: B \to C$, and $f: C \to D$.

• For $(f \circ g) \circ h$: $h: A \to B$, $f \circ g: B \to D$. The composition $(f \circ g) \circ h$ is defined on A and maps to D. $((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = f(g(h(x)))$.
• For $f \circ (g \circ h)$: $g \circ h: A \to C$, $f: C \to D$. The composition $f \circ (g \circ h)$ is defined on A and maps to D. $(f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(g(h(x)))$.

In both cases, the result is the same function mapping from A to D, defined by $x \mapsto f(g(h(x)))$.

Therefore, the composition of functions is indeed associative.

Answer:

True

Statement:

Every function is invertible.

Analysis:

A function is invertible if and only if it is bijective.

A function $f: A \to B$ is bijective if it is both injective (one-to-one) and surjective (onto).

Injective (One-to-one): A function $f$ is injective if distinct elements in the domain map to distinct elements in the codomain. That is, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Surjective (Onto): A function $f$ is surjective if its range is equal to its codomain. That is, for every element $y$ in the codomain, there exists at least one element $x$ in the domain such that $f(x) = y$.

For a function to be invertible, it must satisfy both conditions. Many functions do not satisfy one or both of these conditions.

Consider a counterexample:

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^2$. The domain is the set of real numbers $\mathbb{R}$, and the codomain is also the set of real numbers $\mathbb{R}$.

Let's check if this function is injective:

Consider $x_1 = 2$ and $x_2 = -2$. Both $x_1, x_2 \in \mathbb{R}$.

$f(2) = 2^2 = 4$

$f(-2) = (-2)^2 = 4$

Here, $f(2) = f(-2) = 4$, but $2 \neq -2$. Since different inputs can produce the same output, the function $f(x) = x^2$ is not injective on $\mathbb{R}$.

Let's check if this function is surjective:

The codomain is $\mathbb{R}$. The output of $f(x) = x^2$ for any real number $x$ is always non-negative. The range of $f(x) = x^2$ is $[0, \infty)$.

The range $[0, \infty)$ is not equal to the codomain $\mathbb{R}$. For example, there is no real number $x$ such that $f(x) = -5$ (since $x^2 = -5$ has no real solution). So, negative numbers in the codomain do not have preimages in the domain.

Since the range is not equal to the codomain, the function $f(x) = x^2$ is not surjective on $\mathbb{R}$.

Since the function $f(x) = x^2$ is neither injective nor surjective (and thus not bijective) on the domain and codomain $\mathbb{R}$, it is not invertible.

This counterexample shows that there exists at least one function that is not invertible. Therefore, the statement "Every function is invertible" is false.

Answer:

False

Statement:

A binary operation on a set has always the identity element.

Analysis:

A binary operation $*$ on a set $S$ is a function that takes two elements from $S$ and returns a single element in $S$. Formally, $*: S \times S \to S$.

An identity element $e$ for a binary operation $*$ on a set $S$ is an element $e \in S$ such that for all elements $a \in S$, the following two conditions hold:

For $e$ to be the identity element, it must satisfy both properties for all $a \in S$. Such an element, if it exists, is unique.

The statement claims that *every* binary operation on *any* set has an identity element. To prove this statement false, we need to find just one example of a set and a binary operation on that set that does not have an identity element.

Consider the set of integers, $\mathbb{Z}$, and the binary operation of subtraction, denoted by $-$. Subtraction is a binary operation on $\mathbb{Z}$ because for any two integers $a, b \in \mathbb{Z}$, $a - b$ is also a unique integer in $\mathbb{Z}$.

Now, let's check if there is an identity element $e \in \mathbb{Z}$ for subtraction.

For $e$ to be a right identity, it must satisfy $a - e = a$ for all $a \in \mathbb{Z}$.

From $a - e = a$, we can subtract $a$ from both sides to get $-e = 0$, which means $e = 0$.

So, if a right identity exists, it must be 0. Let's check if 0 is indeed a right identity for subtraction on $\mathbb{Z}$. For any $a \in \mathbb{Z}$, $a - 0 = a$. This is true for all integers $a$. So, 0 is a right identity.

For $e$ to be a left identity, it must satisfy $e - a = a$ for all $a \in \mathbb{Z}$.

Substituting $e=0$ (since an identity element must be unique if it exists, and we found the candidate for the right identity), we must check if $0 - a = a$ for all $a \in \mathbb{Z}$.

The equation $0 - a = a$ simplifies to $-a = a$. This equation is only true when $a = 0$ (since $-a = a \implies 2a = 0 \implies a = 0$).

This property ($0 - a = a$) does not hold for all integers $a$. For example, if $a=5$, $0 - 5 = -5$, and $-5 \neq 5$. If $a=-3$, $0 - (-3) = 3$, and $3 \neq -3$.

Since $e=0$ satisfies the right identity property but not the left identity property for all $a \in \mathbb{Z}$, and there is no other element that could be the identity (as the right identity is unique if it exists), the binary operation of subtraction on $\mathbb{Z}$ does not have a two-sided identity element.

Since we found a binary operation (subtraction) on a set ($\mathbb{Z}$) that does not have an identity element, the statement "A binary operation on a set has always the identity element" is false.

Answer:

False