Chapter 2 Relations And Functions

Given:

The equality of two ordered pairs:

$(x + 1, y – 2) = (3, 1)$

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To Find:

The values of $x$ and $y$.

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Solution:

Two ordered pairs are equal if and only if their corresponding components are equal.

From the given equality, we can equate the first components and the second components separately.

Equating the first components:

$x + 1 = 3$

To find $x$, subtract 1 from both sides of the equation:

$x = 3 - 1$

$x = 2$

Equating the second components:

$y – 2 = 1$

To find $y$, add 2 to both sides of the equation:

$y = 1 + 2$

$y = 3$

Therefore, the values of $x$ and $y$ are $x = 2$ and $y = 3$.

Given:

Set P = ${a, b, c}$

Set Q = ${r}$

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To Find:

The sets P $\times$ Q and Q $\times$ P.

Check if P $\times$ Q = Q $\times$ P.

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Solution:

The Cartesian product P $\times$ Q is the set of all ordered pairs $(p, q)$ where $p \in P$ and $q \in Q$.

P $\times$ Q = ${(a, r), (b, r), (c, r)}$

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The Cartesian product Q $\times$ P is the set of all ordered pairs $(q, p)$ where $q \in Q$ and $p \in P$.

Q $\times$ P = ${(r, a), (r, b), (r, c)}$

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Comparison:

Comparing the elements of P $\times$ Q and Q $\times$ P, we observe that the ordered pairs are different.

For example, the ordered pair $(a, r)$ is an element of P $\times$ Q, but it is not an element of Q $\times$ P.

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Conclusion:

Since the sets P $\times$ Q and Q $\times$ P do not contain the same ordered pairs, they are not equal.

Therefore, P $\times$ Q $\neq$ Q $\times$ P.

Given:

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

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

Set C = $\{4, 5, 6\}$

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To Find:

(i) A $\times$ (B $\cap$ C)

(ii) (A $\times$ B) $\cap$ (A $\times$ C)

(iii) A $\times$ (B $\cup$ C)

(iv) (A $\times$ B) $\cup$ (A $\times$ C)

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Solution:

First, let's find the required set operations on B and C.

Intersection of B and C:

B $\cap$ C = $\{3, 4\} \cap \{4, 5, 6\}$

B $\cap$ C = $\{4\}$

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Union of B and C:

B $\cup$ C = $\{3, 4\} \cup \{4, 5, 6\}$

B $\cup$ C = $\{3, 4, 5, 6\}$

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Now we can find the Cartesian products.

(i) A $\times$ (B $\cap$ C):

This is the Cartesian product of set A and the set (B $\cap$ C).

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

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

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(ii) (A $\times$ B) $\cap$ (A $\times$ C):

First, find A $\times$ B:

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

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

Next, find A $\times$ C:

A $\times$ C = $\{1, 2, 3\} \times \{4, 5, 6\}$

A $\times$ C = $\{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$

Now, find the intersection of A $\times$ B and A $\times$ C:

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

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(iii) A $\times$ (B $\cup$ C):

This is the Cartesian product of set A and the set (B $\cup$ C).

A $\times$ (B $\cup$ C) = $\{1, 2, 3\} \times \{3, 4, 5, 6\}$

A $\times$ (B $\cup$ C) = $\{(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)\}$

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(iv) (A $\times$ B) $\cup$ (A $\times$ C):

We already found A $\times$ B and A $\times$ C in part (ii).

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

A $\times$ C = $\{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$

Now, find the union of A $\times$ B and A $\times$ C:

(A $\times$ B) $\cup$ (A $\times$ C) = $\{(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)\}$

Given:

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

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To Find:

The set P $\times$ P $\times$ P.

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Solution:

The set P $\times$ P $\times$ P represents the set of all ordered triplets $(a, b, c)$ where $a \in P$, $b \in P$, and $c \in P$.

First, let's find P $\times$ P:

P $\times$ P = $\{1, 2\} \times \{1, 2\}$

P $\times$ P = $\{(1, 1), (1, 2), (2, 1), (2, 2)\}$

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Now, we find P $\times$ P $\times$ P, which is the Cartesian product of (P $\times$ P) and P:

P $\times$ P $\times$ P = (P $\times$ P) $\times$ P

P $\times$ P $\times$ P = $\{(1, 1), (1, 2), (2, 1), (2, 2)\} \times \{1, 2\}$

To form the ordered triplets, we take each ordered pair from P $\times$ P and combine it with each element from P.

For the ordered pair $(1, 1)$ from P $\times$ P, combined with elements from P $\{1, 2\}$:

$(1, 1, 1)$ and $(1, 1, 2)$

For the ordered pair $(1, 2)$ from P $\times$ P, combined with elements from P $\{1, 2\}$:

$(1, 2, 1)$ and $(1, 2, 2)$

For the ordered pair $(2, 1)$ from P $\times$ P, combined with elements from P $\{1, 2\}$:

$(2, 1, 1)$ and $(2, 1, 2)$

For the ordered pair $(2, 2)$ from P $\times$ P, combined with elements from P $\{1, 2\}$:

$(2, 2, 1)$ and $(2, 2, 2)$

Combining all these ordered triplets, we get the set P $\times$ P $\times$ P:

P $\times$ P $\times$ P = $\{(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)\}$

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Result:

P $\times$ P $\times$ P = $\{(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)\}$

Given:

R is the set of all real numbers.

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To Represent:

The Cartesian products R $\times$ R and R $\times$ R $\times$ R.

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Solution:

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

Similarly, the Cartesian product of three sets A, B, and C, denoted by A $\times$ B $\times$ C, is the set of all ordered triplets $(a, b, c)$ where $a \in A$, $b \in B$, and $c \in C$.

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Representation of R $\times$ R:

R $\times$ R is the set of all ordered pairs $(x, y)$ where $x \in R$ and $y \in R$.

Each ordered pair $(x, y)$ corresponds to a unique point in a two-dimensional coordinate system (also known as the Cartesian plane).

Therefore, R $\times$ R represents the set of all points in the two-dimensional plane.

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Representation of R $\times$ R $\times$ R:

R $\times$ R $\times$ R is the set of all ordered triplets $(x, y, z)$ where $x \in R$, $y \in R$, and $z \in R$.

Each ordered triplet $(x, y, z)$ corresponds to a unique point in a three-dimensional coordinate system (also known as the Cartesian space).

Therefore, R $\times$ R $\times$ R represents the set of all points in three-dimensional space.

Given:

A $\times$ B = $\{(p, q), (p, r), (m, q), (m, r)\}$.

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To Find:

The sets A and B.

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Solution:

We know that the Cartesian product A $\times$ B is defined as the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$.

A $\times$ B = $\{(a, b) : a \in A, b \in B\}$

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From the given A $\times$ B = $\{(p, q), (p, r), (m, q), (m, r)\}$, the first element of each ordered pair belongs to set A, and the second element of each ordered pair belongs to set B.

The first elements in the ordered pairs are p, p, m, m.

Therefore, the elements of set A are the unique first elements: $A = \{p, m\}$.

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The second elements in the ordered pairs are q, r, q, r.

Therefore, the elements of set B are the unique second elements: $B = \{q, r\}$.

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Result:

The sets A and B are:

A = $\{p, m\}$

B = $\{q, r\}$

Given:

$\left( \frac{x}{3}+1 , \;y \;-\; \frac{2}{3} \right) = \left( \frac{5}{3},\frac{1}{3} \right)$

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To Find:

The values of x and y.

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Solution:

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

Therefore, we can equate the first components and the second components separately.

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Equating the first components:

$\frac{x}{3} + 1 = \frac{5}{3}$

Subtract 1 from both sides:

$\frac{x}{3} = \frac{5}{3} - 1$

Convert 1 to a fraction with denominator 3:

$\frac{x}{3} = \frac{5}{3} - \frac{3}{3}$

Simplify the right side:

$\frac{x}{3} = \frac{5 - 3}{3}$

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

Multiply both sides by 3:

$x = \frac{2}{3} \times 3$

$x = 2$

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Equating the second components:

$y - \frac{2}{3} = \frac{1}{3}$

Add $\frac{2}{3}$ to both sides:

$y = \frac{1}{3} + \frac{2}{3}$

Combine the fractions:

$y = \frac{1 + 2}{3}$

$y = \frac{3}{3}$

Simplify:

$y = 1$

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Result:

The values of x and y are:

$x = 2$ and $y = 1$.

Given:

The number of elements in set A is 3. So, $|A| = 3$.

The set B = $\{3, 4, 5\}$.

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To Find:

The number of elements in (A $\times$ B), i.e., $|A \times B|$.

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Solution:

We know that if A and B are finite sets, then the number of elements in the Cartesian product A $\times$ B is the product of the number of elements in A and the number of elements in B.

The formula is given by:

$|A \times B| = |A| \times |B|$

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From the given information, we have:

$|A| = 3$

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

The number of elements in set B is 3. So, $|B| = 3$.

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Now, using the formula for the number of elements in the Cartesian product:

$|A \times B| = |A| \times |B|$

$|A \times B| = 3 \times 3$

$|A \times B| = 9$

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Result:

The number of elements in (A $\times$ B) is 9.

Given:

Set G = $\{7, 8\}$

Set H = $\{5, 4, 2\}$

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To Find:

The Cartesian products G $\times$ H and H $\times$ G.

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Solution:

The Cartesian product G $\times$ H is the set of all ordered pairs $(g, h)$ where $g \in G$ and $h \in H$.

G $\times$ H = $\{(g, h) : g \in \{7, 8\}, h \in \{5, 4, 2\}\}$

We list all possible ordered pairs:

G $\times$ H = $\{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$

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The Cartesian product H $\times$ G is the set of all ordered pairs $(h, g)$ where $h \in H$ and $g \in G$.

H $\times$ G = $\{(h, g) : h \in \{5, 4, 2\}, g \in \{7, 8\}\}$

We list all possible ordered pairs:

H $\times$ G = $\{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$

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Result:

G $\times$ H = $\{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$

H $\times$ G = $\{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$

To Determine:

Whether each statement is true or false, and rewrite false statements correctly.

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Solution:

(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.

Let's find the Cartesian product P $\times$ Q.

P $\times$ Q = $\{(p, q) : p \in P, q \in Q\}$

P $\times$ Q = $\{(m, n), (m, m), (n, n), (n, m)\}$

The given statement says P $\times$ Q = $\{(m, n), (n, m)\}$. This is not equal to the actual P $\times$ Q calculated above.

Therefore, the statement is False.

The correct statement is:

If P = $\{m, n\}$ and Q = $\{n, m\}$, then P $\times$ Q = $\{(m, n), (m, m), (n, n), (n, m)\}$.

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(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.

The definition of the Cartesian product A $\times$ B is the set of all ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$.

If A is non-empty, it contains at least one element. If B is non-empty, it contains at least one element.

If $a \in A$ and $b \in B$, then the ordered pair $(a, b)$ is an element of A $\times$ B. Since both A and B are non-empty, there exists at least one such ordered pair $(a, b)$.

Therefore, A $\times$ B is a non-empty set.

The statement accurately describes the Cartesian product of two non-empty sets.

Therefore, the statement is True.

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(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

First, let's find the intersection of set B and the empty set $\phi$ (phi).

B $\cap$ $\phi$ = $\{3, 4\} \cap \phi$

The intersection of any set with the empty set is the empty set.

B $\cap$ $\phi$ = $\phi$

Now, we need to find the Cartesian product of set A and the result of the intersection:

A $\times$ (B $\cap$ $\phi$) = A $\times$ $\phi$

A $\times$ $\phi$ = $\{1, 2\} \times \phi$

The Cartesian product of a non-empty set with the empty set is the empty set.

A $\times$ $\phi$ = $\phi$

The given statement A $\times$ (B $\cap$ $\phi$) = $\phi$ is consistent with our calculation.

Therefore, the statement is True.

Given:

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

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To Find:

The set A $\times$ A $\times$ A.

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Solution:

The set A $\times$ A $\times$ A represents the set of all ordered triplets $(a, b, c)$ where $a \in A$, $b \in A$, and $c \in A$.

First, let's find A $\times$ A:

A $\times$ A = $\{-1, 1\} \times \{-1, 1\}$

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

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Now, we find A $\times$ A $\times$ A, which is the Cartesian product of (A $\times$ A) and A:

A $\times$ A $\times$ A = (A $\times$ A) $\times$ A

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

To form the ordered triplets, we take each ordered pair from A $\times$ A and combine it with each element from A.

• From $(-1, -1)$ and $\{-1, 1\}$: $(-1, -1, -1), (-1, -1, 1)$
• From $(-1, 1)$ and $\{-1, 1\}$: $(-1, 1, -1), (-1, 1, 1)$
• From $(1, -1)$ and $\{-1, 1\}$: $(1, -1, -1), (1, -1, 1)$
• From $(1, 1)$ and $\{-1, 1\}$: $(1, 1, -1), (1, 1, 1)$

Combining all these ordered triplets, we get the set A $\times$ A $\times$ A:

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

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Result:

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

Given:

A $\times$ B = $\{(a, x), (a, y), (b, x), (b, y)\}$.

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To Find:

The sets A and B.

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Solution:

We know that the Cartesian product A $\times$ B is the set of all ordered pairs $(u, v)$ where $u \in A$ and $v \in B$.

A $\times$ B = $\{(u, v) : u \in A, v \in B\}$

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From the given set A $\times$ B, the first component of each ordered pair is an element of set A.

The first components are a, a, b, and b.

The set A consists of the unique values of the first components.

So, A = $\{a, b\}$.

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From the given set A $\times$ B, the second component of each ordered pair is an element of set B.

The second components are x, y, x, and y.

The set B consists of the unique values of the second components.

So, B = $\{x, y\}$.

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Result:

The sets A and B are:

A = $\{a, b\}$

B = $\{x, y\}$

Given:

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

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

Set C = $\{5, 6\}$

Set D = $\{5, 6, 7, 8\}$

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To Verify:

(i) A $\times$ (B $\cap$ C) = (A $\times$ B) $\cap$ (A $\times$ C)

(ii) A $\times$ C is a subset of B $\times$ D

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Verification of (i):

First, let's find B $\cap$ C:

B $\cap$ C = $\{1, 2, 3, 4\} \cap \{5, 6\}$

Since there are no common elements between B and C, the intersection is the empty set $\phi$.

B $\cap$ C = $\phi$

Now, let's find A $\times$ (B $\cap$ C):

A $\times$ (B $\cap$ C) = A $\times$ $\phi$

A $\times$ (B $\cap$ C) = $\{1, 2\} \times \phi$

The Cartesian product of any set with the empty set is the empty set.

A $\times$ (B $\cap$ C) = $\phi$

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Next, let's find A $\times$ B and A $\times$ C.

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

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

A $\times$ C = $\{1, 2\} \times \{5, 6\}$

A $\times$ C = $\{(1, 5), (1, 6), (2, 5), (2, 6)\}$

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Now, let's find the intersection of A $\times$ B and A $\times$ C:

(A $\times$ B) $\cap$ (A $\times$ C) = $\{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)\} \cap \{(1, 5), (1, 6), (2, 5), (2, 6)\}$

There are no common ordered pairs in these two sets.

(A $\times$ B) $\cap$ (A $\times$ C) = $\phi$

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Comparing A $\times$ (B $\cap$ C) = $\phi$ and (A $\times$ B) $\cap$ (A $\times$ C) = $\phi$.

Since both results are the empty set, the statement is true.

A $\times$ (B $\cap$ C) = (A $\times$ B) $\cap$ (A $\times$ C)

Statement (i) is Verified.

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Verification of (ii):

To verify if A $\times$ C is a subset of B $\times$ D, we need to show that every element in A $\times$ C is also an element in B $\times$ D.

First, let's find A $\times$ C:

A $\times$ C = $\{1, 2\} \times \{5, 6\}$

A $\times$ C = $\{(1, 5), (1, 6), (2, 5), (2, 6)\}

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Next, let's find B $\times$ D:

B $\times$ D = $\{1, 2, 3, 4\} \times \{5, 6, 7, 8\}$

B $\times$ D = $\{(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)\}

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Now, check if every element of A $\times$ C is in B $\times$ D:

• Is (1, 5) in B $\times$ D? Yes, because $1 \in B$ and $5 \in D$.
• Is (1, 6) in B $\times$ D? Yes, because $1 \in B$ and $6 \in D$.
• Is (2, 5) in B $\times$ D? Yes, because $2 \in B$ and $5 \in D$.
• Is (2, 6) in B $\times$ D? Yes, because $2 \in B$ and $6 \in D$.

Since every element in A $\times$ C is also present in B $\times$ D, A $\times$ C is a subset of B $\times$ D.

A $\times$ C $\subseteq$ B $\times$ D

Statement (ii) is Verified.

Given:

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

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

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To Find:

The set A $\times$ B.

The number of subsets of A $\times$ B.

List all subsets of A $\times$ B.

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Solution:

First, we find the Cartesian product A $\times$ B.

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

A $\times$ B = $\{(a, b) : a \in \{1, 2\}, b \in \{3, 4\}\}$

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

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Next, we find the number of elements in A $\times$ B.

The number of elements in set A is $|A| = 2$.

The number of elements in set B is $|B| = 2$.

The number of elements in A $\times$ B is $|A \times B| = |A| \times |B| = 2 \times 2 = 4$.

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

Here, the number of elements in A $\times$ B is 4.

Number of subsets of A $\times$ B = $2^{|A \times B|} = 2^4 = 16$.

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Now, we list all the 16 subsets of A $\times$ B = $\{(1, 3), (1, 4), (2, 3), (2, 4)\}$.

Let the elements of A $\times$ B be $e_1 = (1, 3)$, $e_2 = (1, 4)$, $e_3 = (2, 3)$, and $e_4 = (2, 4)$.

Subsets with 0 elements (the empty set):

$\phi$

Subsets with 1 element:

$\{(1, 3)\}$

$\{(1, 4)\}$

$\{(2, 3)\}$

$\{(2, 4)\}$

Subsets with 2 elements:

$\{(1, 3), (1, 4)\}$

$\{(1, 3), (2, 3)\}$

$\{(1, 3), (2, 4)\}$

$\{(1, 4), (2, 3)\}$

$\{(1, 4), (2, 4)\}$

$\{(2, 3), (2, 4)\}$

Subsets with 3 elements:

$\{(1, 3), (1, 4), (2, 3)\}$

$\{(1, 3), (1, 4), (2, 4)\}$

$\{(1, 3), (2, 3), (2, 4)\}$

$\{(1, 4), (2, 3), (2, 4)\}$

Subsets with 4 elements (the set itself):

$\{(1, 3), (1, 4), (2, 3), (2, 4)\}$

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Result:

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

The number of subsets of A $\times$ B is 16.

The subsets are: $\phi$, $\{(1, 3)\}$, $\{(1, 4)\}$, $\{(2, 3)\}$, $\{(2, 4)\}$, $\{(1, 3), (1, 4)\}$, $\{(1, 3), (2, 3)\}$, $\{(1, 3), (2, 4)\}$, $\{(1, 4), (2, 3)\}$, $\{(1, 4), (2, 4)\}$, $\{(2, 3), (2, 4)\}$, $\{(1, 3), (1, 4), (2, 3)\}$, $\{(1, 3), (1, 4), (2, 4)\}$, $\{(1, 3), (2, 3), (2, 4)\}$, $\{(1, 4), (2, 3), (2, 4)\}$, and $\{(1, 3), (1, 4), (2, 3), (2, 4)\}$.

Given:

n(A) = 3

n(B) = 2

The ordered pairs $(x, 1)$, $(y, 2)$, and $(z, 1)$ are in A $\times$ B, where x, y, and z are distinct elements.

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To Find:

The sets A and B.

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Solution:

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

From the definition, the first component of every ordered pair in A $\times$ B must belong to set A.

The given ordered pairs in A $\times$ B are $(x, 1)$, $(y, 2)$, and $(z, 1)$.

The first components are x, y, and z.

Since these first components x, y, and z are given to be distinct elements, and we know that n(A) = 3, these three elements must be the elements of set A.

Therefore, A = $\{x, y, z\}$.

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The second component of every ordered pair in A $\times$ B must belong to set B.

The second components are 1, 2, and 1.

The unique second components are 1 and 2.

We are given that n(B) = 2, which means set B has exactly two elements.

Therefore, the elements of set B must be 1 and 2.

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

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Result:

The sets A and B are:

A = $\{x, y, z\}$

B = $\{1, 2\}$

Given:

The Cartesian product A $\times$ A has 9 elements, i.e., $|A \times A| = 9$.

Among the elements of A $\times$ A are $(-1, 0)$ and $(0, 1)$.

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To Find:

The set A.

The remaining elements of A $\times$ A.

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Solution:

We know that for any finite set A, the number of elements in the Cartesian product A $\times$ A is given by $|A \times A| = |A| \times |A| = |A|^2$.

Given that $|A \times A| = 9$, we have:

$|A|^2 = 9$

$|A| = \sqrt{9}$

Since the number of elements must be non-negative, $|A| = 3$.

This means that set A has 3 elements.

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The elements of A $\times$ A are ordered pairs $(a_1, a_2)$, where $a_1 \in A$ and $a_2 \in A$.

We are given that $(-1, 0) \in A \times A$. This implies that the first component, $-1$, must be in A, and the second component, $0$, must also be in A.

So, $-1 \in A$ and $0 \in A$.

We are also given that $(0, 1) \in A \times A$. This implies that the first component, $0$, must be in A, and the second component, $1$, must also be in A.

So, $0 \in A$ and $1 \in A$.

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Combining the elements we've found that must be in A, we have $\{-1, 0, 1\}$.

Since we know that set A has exactly 3 elements, and we have identified three distinct elements $(-1, 0, 1)$ that belong to A, these must be all the elements of set A.

Therefore, the set A = $\{-1, 0, 1\}$.

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Now, we find the complete set A $\times$ A using A = $\{-1, 0, 1\}$.

A $\times$ A = $\{-1, 0, 1\} \times \{-1, 0, 1\}$

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

---

The elements given in the problem are $(-1, 0)$ and $(0, 1)$.

The remaining elements of A $\times$ A are the elements in the complete list that were not given.

The remaining elements are:

$(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)$.

---

Result:

The set A = $\{-1, 0, 1\}$.

The remaining elements of A $\times$ A are $(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)$.

Given:

Set A = $\{1, 2, 3, 4, 5, 6\}$

Relation R from A to A defined by R = $\{(x, y) : y = x + 1\}$

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To Find:

(i) Depict the relation using an arrow diagram.

(ii) Write down the domain, codomain, and range of R.

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Solution:

The relation R consists of ordered pairs $(x, y)$ such that $x \in A$, $y \in A$, and $y = x + 1$.

Let's find the ordered pairs that satisfy the condition $y = x + 1$ for $x, y \in \{1, 2, 3, 4, 5, 6\}$.

• If $x = 1$, then $y = 1 + 1 = 2$. Since $2 \in A$, $(1, 2) \in R$.
• If $x = 2$, then $y = 2 + 1 = 3$. Since $3 \in A$, $(2, 3) \in R$.
• If $x = 3$, then $y = 3 + 1 = 4$. Since $4 \in A$, $(3, 4) \in R$.
• If $x = 4$, then $y = 4 + 1 = 5$. Since $5 \in A$, $(4, 5) \in R$.
• If $x = 5$, then $y = 5 + 1 = 6$. Since $6 \in A$, $(5, 6) \in R$.
• If $x = 6$, then $y = 6 + 1 = 7$. Since $7 \notin A$, $(6, 7) \notin R$.

So, the relation R as a set of ordered pairs is:

R = $\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$.

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(i) Arrow Diagram:

To depict the relation using an arrow diagram, we draw two sets (representing A) and draw arrows from elements of the first set to corresponding elements in the second set based on the ordered pairs in R.

We represent the set A elements in two columns and draw arrows as follows:

        1  ----->  2
        2  ----->  3
        3  ----->  4
        4  ----->  5
        5  ----->  6
        6
        

(Note: HTML pre tag is used for simple text diagrammatic representation as complex graphical diagrams are not feasible directly in HTML with the given constraints. A visual arrow diagram would typically show two ovals representing set A, with points inside, and arrows drawn between points according to the relation R).

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(ii) Domain, Codomain, and Range of R:

The domain of a relation R from set A to set B is the set of all first components of the ordered pairs in R.

From R = $\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$, the first components are 1, 2, 3, 4, and 5.

Domain of R = $\{1, 2, 3, 4, 5\}$.

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The codomain of a relation R from set A to set A is the set A itself.

Codomain of R = A = $\{1, 2, 3, 4, 5, 6\}$.

---

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

From R = $\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$, the second components are 2, 3, 4, 5, and 6.

Range of R = $\{2, 3, 4, 5, 6\}$.

---

Result:

(i) Arrow diagram is depicted above (text-based). A visual diagram involves drawing elements of A and arrows.

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

Codomain of R = $\{1, 2, 3, 4, 5, 6\}$

Range of R = $\{2, 3, 4, 5, 6\}$

Given:

Arrow diagram showing a relation from set P to set Q.

Set P = $\{5, 6, 7\}$

Set Q = $\{3, 4, 5\}$

The relation consists of the ordered pairs indicated by the arrows:

• From 5 in P to 3 in Q, giving the ordered pair (5, 3).
• From 6 in P to 4 in Q, giving the ordered pair (6, 4).
• From 7 in P to 5 in Q, giving the ordered pair (7, 5).

---

To Find:

(i) Relation in set-builder form.

(ii) Relation in roster form.

(iii) Domain and range of the relation.

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Solution:

Let R be the relation from P to Q.

(ii) Roster Form:

The roster form lists all the ordered pairs in the relation.

R = $\{(5, 3), (6, 4), (7, 5)\}$

---

(i) Set-Builder Form:

We need to find a rule that relates the first component ($x$) to the second component ($y$) for each ordered pair $(x, y)$ in the relation.

Looking at the pairs:

• (5, 3): $5 - 2 = 3$
• (6, 4): $6 - 2 = 4$
• (7, 5): $7 - 2 = 5$

The relationship appears to be $y = x - 2$. Also, the first component $x$ belongs to set P, and the second component $y$ belongs to set Q.

In set-builder form, the relation is written as:

R = $\{(x, y) : y = x - 2, x \in P, y \in Q\}$

---

Domain and Range:

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

From R = $\{(5, 3), (6, 4), (7, 5)\}$, the first components are 5, 6, and 7.

Domain of R = $\{5, 6, 7\}$

---

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

From R = $\{(5, 3), (6, 4), (7, 5)\}$, the second components are 3, 4, and 5.

Range of R = $\{3, 4, 5\}$

---

Result:

(i) Set-builder form: R = $\{(x, y) : y = x - 2, x \in P, y \in Q\}$

(ii) Roster form: R = $\{(5, 3), (6, 4), (7, 5)\}$

(iii) Domain of R = $\{5, 6, 7\}$

Range of R = $\{3, 4, 5\}$

Given:

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

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

---

To Find:

The number of relations from A to B.

---

Solution:

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

---

First, let's find the Cartesian product A $\times$ B.

A $\times$ B = $\{(a, b) : a \in A, b \in B\}$

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

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

---

Next, we find the number of elements in the set A $\times$ B.

The number of elements in set A is $|A| = 2$.

The number of elements in set B is $|B| = 2$.

The number of elements in A $\times$ B is $|A \times B| = |A| \times |B| = 2 \times 2 = 4$.

---

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

Since a relation from A to B is a subset of A $\times$ B, the number of possible relations is equal to the number of subsets of A $\times$ B.

Number of relations from A to B = Number of subsets of (A $\times$ B)

Number of relations = $2^{|A \times B|}$

Number of relations = $2^4$

$2^4 = 2 \times 2 \times 2 \times 2 = 16$

---

Result:

The number of relations from A to B is 16.

Given:

Set A = $\{1, 2, 3, ..., 14\}$

Relation R from A to A defined by R = $\{(x, y) : 3x - y = 0, \text{ where } x, y \in A\}$

---

To Find:

The domain, codomain, and range of R.

---

Solution:

The relation is given by the condition $3x - y = 0$, which can be rewritten as $y = 3x$.

We need to find all ordered pairs $(x, y)$ such that $x \in A$, $y \in A$, and $y = 3x$.

Let's find the values of y for each possible value of x from set A:

• If $x = 1$, $y = 3(1) = 3$. Since $3 \in A$, $(1, 3)$ is in R.
• If $x = 2$, $y = 3(2) = 6$. Since $6 \in A$, $(2, 6)$ is in R.
• If $x = 3$, $y = 3(3) = 9$. Since $9 \in A$, $(3, 9)$ is in R.
• If $x = 4$, $y = 3(4) = 12$. Since $12 \in A$, $(4, 12)$ is in R.
• If $x = 5$, $y = 3(5) = 15$. Since $15 \notin A$ (as A goes only up to 14), $(5, 15)$ is not in R.

For any value of x greater than 4 in set A, the value of y ($3x$) will be greater than 14, and thus not in set A.

So, the relation R in roster form is:

R = $\{(1, 3), (2, 6), (3, 9), (4, 12)\}$

---

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

Domain of R = $\{1, 2, 3, 4\}$

---

The codomain of a relation from set A to set A is the set A itself.

Codomain of R = A = $\{1, 2, 3, ..., 14\}$

---

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

Range of R = $\{3, 6, 9, 12\}$

---

Result:

Domain of R = $\{1, 2, 3, 4\}$

Codomain of R = $\{1, 2, 3, ..., 14\}$

Range of R = $\{3, 6, 9, 12\}$

Given:

Relation R on the set N of natural numbers.

R = $\{(x, y) : y = x + 5, \text{ x is a natural number less than 4; } x, y \in N\}$.

---

To Find:

The relation in roster form.

The domain of R.

The range of R.

---

Solution:

The set of natural numbers N is $\{1, 2, 3, 4, ...\}$.

The relation R is defined for $x \in N$ such that $x$ is less than 4. So, the possible values for $x$ are $1, 2, 3$.

The condition relating x and y is $y = x + 5$, and both x and y must be natural numbers.

---

Let's find the ordered pairs $(x, y)$ for the possible values of x:

• If $x = 1$, then $y = 1 + 5 = 6$. Since $y = 6 \in N$, the ordered pair is $(1, 6)$.
• If $x = 2$, then $y = 2 + 5 = 7$. Since $y = 7 \in N$, the ordered pair is $(2, 7)$.
• If $x = 3$, then $y = 3 + 5 = 8$. Since $y = 8 \in N$, the ordered pair is $(3, 8)$.

These are all the ordered pairs that satisfy the conditions of the relation.

---

Roster Form:

The relation R in roster form is the set of these ordered pairs:

R = $\{(1, 6), (2, 7), (3, 8)\}$.

---

Domain of R:

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

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

---

Range of R:

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

Range(R) = $\{6, 7, 8\}$.

---

Result:

The relation R in roster form is R = $\{(1, 6), (2, 7), (3, 8)\}$.

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

The range of R is $\{6, 7, 8\}$.

Given:

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

Set B = $\{4, 6, 9\}$

Relation R from A to B defined by R = $\{(x, y) : \text{the difference between } x \text{ and } y \text{ is odd; } x \in A, y \in B\}$

---

To Find:

The relation R in roster form.

---

Solution:

The condition "the difference between x and y is odd" means that $|x - y|$ is an odd number. This occurs when one of x and y is odd and the other is even.

Let's check the parity of the elements in sets A and B.

Set A = $\{1 \text{ (odd)}, 2 \text{ (even)}, 3 \text{ (odd)}, 5 \text{ (odd)}\}$.

Set B = $\{4 \text{ (even)}, 6 \text{ (even)}, 9 \text{ (odd)}\}$.

---

We need to find ordered pairs $(x, y)$ such that $x \in A$, $y \in B$, and one of x or y is odd while the other is even.

Consider $x \in A$ and check the pairs $(x, y)$ for all $y \in B$ based on the parity rule:

• For $x = 1$ (odd): We need $y \in B$ to be even. The even numbers in B are 4 and 6.
  • $(1, 4)$: Difference $|1 - 4| = |-3| = 3$, which is odd. $(1, 4) \in R$.
  • $(1, 6)$: Difference $|1 - 6| = |-5| = 5$, which is odd. $(1, 6) \in R$.
  • $(1, 9)$: Difference $|1 - 9| = |-8| = 8$, which is even. $(1, 9) \notin R$.
• For $x = 2$ (even): We need $y \in B$ to be odd. The odd number in B is 9.
  • $(2, 4)$: Difference $|2 - 4| = |-2| = 2$, which is even. $(2, 4) \notin R$.
  • $(2, 6)$: Difference $|2 - 6| = |-4| = 4$, which is even. $(2, 6) \notin R$.
  • $(2, 9)$: Difference $|2 - 9| = |-7| = 7$, which is odd. $(2, 9) \in R$.
• For $x = 3$ (odd): We need $y \in B$ to be even. The even numbers in B are 4 and 6.
  • $(3, 4)$: Difference $|3 - 4| = |-1| = 1$, which is odd. $(3, 4) \in R$.
  • $(3, 6)$: Difference $|3 - 6| = |-3| = 3$, which is odd. $(3, 6) \in R$.
  • $(3, 9)$: Difference $|3 - 9| = |-6| = 6$, which is even. $(3, 9) \notin R$.
• For $x = 5$ (odd): We need $y \in B$ to be even. The even numbers in B are 4 and 6.
  • $(5, 4)$: Difference $|5 - 4| = |1| = 1$, which is odd. $(5, 4) \in R$.
  • $(5, 6)$: Difference $|5 - 6| = |-1| = 1$, which is odd. $(5, 6) \in R$.
  • $(5, 9)$: Difference $|5 - 9| = |-4| = 4$, which is even. $(5, 9) \notin R$.

The ordered pairs that satisfy the condition are the ones we found to be in R.

---

Roster Form:

The relation R in roster form is the set of these ordered pairs:

R = $\{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}$

---

Result:

R = $\{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}$.

Given:

An arrow diagram showing a relation from set P to set Q.

From the diagram, set P = $\{5, 6, 7\}$ and set Q = $\{3, 4, 5\}$.

The arrows represent the following ordered pairs in the relation:

• 5 is related to 3, so $(5, 3)$ is in the relation.
• 6 is related to 4, so $(6, 4)$ is in the relation.
• 7 is related to 5, so $(7, 5)$ is in the relation.

---

To Find:

(i) The relation in set-builder form.

(ii) The relation in roster form.

(iii) The domain and range of the relation.

---

Solution:

Let R be the relation from set P to set Q shown in the diagram.

---

(ii) Roster Form:

The roster form of the relation is the set of all ordered pairs indicated by the arrows:

R = $\{(5, 3), (6, 4), (7, 5)\}$

---

(i) Set-Builder Form:

We look for a pattern or rule relating the first component ($x$) to the second component ($y$) in each ordered pair $(x, y) \in R$.

• For $(5, 3)$, $5 - 2 = 3$.
• For $(6, 4)$, $6 - 2 = 4$.
• For $(7, 5)$, $7 - 2 = 5$.

The rule is $y = x - 2$.

The set-builder form of the relation R is:

R = $\{(x, y) : y = x - 2, x \in P, y \in Q\}$

---

Domain and Range:

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

From R = $\{(5, 3), (6, 4), (7, 5)\}$, the first components are 5, 6, and 7.

Domain of R = $\{5, 6, 7\}$.

---

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

From R = $\{(5, 3), (6, 4), (7, 5)\}$, the second components are 3, 4, and 5.

Range of R = $\{3, 4, 5\}$.

---

Result:

(i) Set-builder form: R = $\{(x, y) : y = x - 2, x \in P, y \in Q\}$

(ii) Roster form: R = $\{(5, 3), (6, 4), (7, 5)\}$

(iii) Domain of R = $\{5, 6, 7\}$

Range of R = $\{3, 4, 5\}$

Given:

Set A = $\{1, 2, 3, 4, 6\}$

Relation R on A defined by R = $\{(a, b) : a, b \in A, b \text{ is exactly divisible by } a\}$

---

To Find:

(i) Relation R in roster form.

(ii) The domain of R.

(iii) The range of R.

---

Solution:

The relation R consists of ordered pairs $(a, b)$ where $a$ and $b$ are elements of set A, and $b$ is exactly divisible by $a$. This means that $\frac{b}{a}$ is an integer.

We need to find all pairs $(a, b)$ from A $\times$ A that satisfy this condition.

---

Let's list the ordered pairs $(a, b)$ such that $a \in A$, $b \in A$, and $b$ is divisible by $a$:

• For $a = 1$: b must be divisible by 1. Elements in A divisible by 1 are 1, 2, 3, 4, 6.

  Ordered pairs: $(1, 1), (1, 2), (1, 3), (1, 4), (1, 6)$

• For $a = 2$: b must be divisible by 2. Elements in A divisible by 2 are 2, 4, 6.

  Ordered pairs: $(2, 2), (2, 4), (2, 6)$

• For $a = 3$: b must be divisible by 3. Elements in A divisible by 3 are 3, 6.

  Ordered pairs: $(3, 3), (3, 6)$

• For $a = 4$: b must be divisible by 4. Elements in A divisible by 4 is 4.

  Ordered pairs: $(4, 4)$

• For $a = 6$: b must be divisible by 6. Elements in A divisible by 6 is 6.

  Ordered pairs: $(6, 6)$

---

(i) Roster Form:

Combining all the ordered pairs found, the relation R in roster form is:

R = $\{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\}$

---

(ii) Domain of R:

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

Domain(R) = $\{1, 2, 3, 4, 6\}$

---

(iii) Range of R:

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

Range(R) = $\{1, 2, 3, 4, 6\}$

---

Result:

(i) R = $\{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\}$

(ii) Domain of R = $\{1, 2, 3, 4, 6\}$

(iii) Range of R = $\{1, 2, 3, 4, 6\}$

Given:

The relation R defined by R = $\{(x, x + 5) : x \in \{0, 1, 2, 3, 4, 5\}\}$.

---

To Find:

The domain of R.

The range of R.

---

Solution:

The relation R consists of ordered pairs $(x, y)$ where $x$ is an element from the set $\{0, 1, 2, 3, 4, 5\}$ and $y = x + 5$.

Let's find the ordered pairs in R by substituting the possible values of $x$:

• If $x = 0$, $y = 0 + 5 = 5$. The ordered pair is $(0, 5)$.
• If $x = 1$, $y = 1 + 5 = 6$. The ordered pair is $(1, 6)$.
• If $x = 2$, $y = 2 + 5 = 7$. The ordered pair is $(2, 7)$.
• If $x = 3$, $y = 3 + 5 = 8$. The ordered pair is $(3, 8)$.
• If $x = 4$, $y = 4 + 5 = 9$. The ordered pair is $(4, 9)$.
• If $x = 5$, $y = 5 + 5 = 10$. The ordered pair is $(5, 10)$.

So, the relation R in roster form is:

R = $\{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$.

---

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

From R = $\{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$, the first components are 0, 1, 2, 3, 4, and 5.

Domain of R = $\{0, 1, 2, 3, 4, 5\}$.

---

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

From R = $\{(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\}$, the second components are 5, 6, 7, 8, 9, and 10.

Range of R = $\{5, 6, 7, 8, 9, 10\}$.

---

Result:

Domain of R = $\{0, 1, 2, 3, 4, 5\}$

Range of R = $\{5, 6, 7, 8, 9, 10\}$

Given:

The relation R defined by R = $\{(x, x^3) : x \text{ is a prime number less than } 10\}$.

---

To Find:

The relation R in roster form.

---

Solution:

The relation R consists of ordered pairs $(x, x^3)$, where the first component $x$ is a prime number less than 10.

First, identify the prime numbers less than 10.

The prime numbers are 2, 3, 5, 7.

---

Now, for each of these prime numbers $x$, we calculate $y = x^3$ to form the ordered pair $(x, y) = (x, x^3)$ that belongs to the relation R.

• If $x = 2$, $y = 2^3 = 2 \times 2 \times 2 = 8$. The ordered pair is $(2, 8)$.
• If $x = 3$, $y = 3^3 = 3 \times 3 \times 3 = 27$. The ordered pair is $(3, 27)$.
• If $x = 5$, $y = 5^3 = 5 \times 5 \times 5 = 125$. The ordered pair is $(5, 125)$.
• If $x = 7$, $y = 7^3 = 7 \times 7 \times 7 = 343$. The ordered pair is $(7, 343)$.

These are all the ordered pairs that satisfy the definition of the relation R.

---

Roster Form:

The relation R in roster form is the set containing these ordered pairs:

R = $\{(2, 8), (3, 27), (5, 125), (7, 343)\}$.

---

Result:

The relation R in roster form is $\{(2, 8), (3, 27), (5, 125), (7, 343)\}$.

Given:

Set A = $\{x, y, z\}$

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

---

To Find:

The number of relations from A to B.

---

Solution:

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

---

First, find the number of elements in set A and set B.

$|A| = 3$

$|B| = 2$

---

Next, find the number of elements in the Cartesian product A $\times$ B.

$|A \times B| = |A| \times |B|$

$|A \times B| = 3 \times 2$

$|A \times B| = 6$

The set A $\times$ B contains 6 ordered pairs.

---

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

Since a relation from A to B is a subset of A $\times$ B, the number of possible relations from A to B is the number of subsets of A $\times$ B.

Number of relations from A to B = $2^{|A \times B|}$

Number of relations = $2^6$

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

---

Result:

The number of relations from A to B is 64.

Given:

Set Z = the set of all integers.

Relation R on Z defined by R = $\{(a, b) : a, b \in Z, a - b \text{ is an integer}\}$.

---

To Find:

The domain of R.

The range of R.

---

Solution:

The relation R consists of ordered pairs $(a, b)$ such that $a$ and $b$ are both integers ($a \in Z, b \in Z$), and the difference $a - b$ is an integer.

We need to determine for which values of $a$ (from Z) there exists some $b$ (from Z) such that $(a, b) \in R$. This will give us the domain.

We also need to determine for which values of $b$ (from Z) there exists some $a$ (from Z) such that $(a, b) \in R$. This will give us the range.

---

Consider any two integers, say $a$ and $b$. The difference $a - b$ is always an integer.

For example, if $a = 5$ and $b = 3$, $a - b = 5 - 3 = 2$, which is an integer.

If $a = -2$ and $b = 7$, $a - b = -2 - 7 = -9$, which is an integer.

If $a = 0$ and $b = -4$, $a - b = 0 - (-4) = 4$, which is an integer.

In general, the difference between any two integers is always an integer. This is a fundamental property of integers.

---

Since for any choice of $a \in Z$ and any choice of $b \in Z$, the condition "$a - b$ is an integer" is always satisfied, every possible ordered pair $(a, b)$ with $a \in Z$ and $b \in Z$ is in the relation R.

Thus, R is the set of all possible ordered pairs of integers, which is the Cartesian product Z $\times$ Z.

R = Z $\times$ Z = $\{(a, b) : a \in Z, b \in Z\}$

---

The domain of R is the set of all first components of the ordered pairs in R. Since the first component $a$ can be any integer from Z, the domain is Z.

Domain(R) = $\{a : (a, b) \in R \text{ for some } b \in Z\}$

Since $(a, b) \in R$ for all $a, b \in Z$, the domain is the set of all integers.

Domain of R = Z

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The range of R is the set of all second components of the ordered pairs in R. Since the second component $b$ can be any integer from Z, the range is Z.

Range(R) = $\{b : (a, b) \in R \text{ for some } a \in Z\}$

Since $(a, b) \in R$ for all $a, b \in Z$, the range is the set of all integers.

Range of R = Z

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Result:

The domain of R is Z.

The range of R is Z.

Given:

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

Relation R is defined on N such that R = $\{(x, y) : y = 2x, x, y \in N\}$.

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To Find:

The domain of R.

The codomain of R.

The range of R.

Whether R is a function.

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Solution:

The relation R consists of all ordered pairs $(x, y)$ where $x$ and $y$ are natural numbers and the second element $y$ is twice the first element $x$.

Let's list some elements of the relation R:

• If $x = 1$, $y = 2(1) = 2$. Since $1 \in N$ and $2 \in N$, $(1, 2) \in R$.
• If $x = 2$, $y = 2(2) = 4$. Since $2 \in N$ and $4 \in N$, $(2, 4) \in R$.
• If $x = 3$, $y = 2(3) = 6$. Since $3 \in N$ and $6 \in N$, $(3, 6) \in R$.
• ...
• If $x = k$ (where $k$ is any natural number), $y = 2k$. Since $k \in N$, $2k$ is also always a natural number ($2k \in N$). So, $(k, 2k) \in R$ for all $k \in N$.

In roster form, the relation can be written as R = $\{(1, 2), (2, 4), (3, 6), (4, 8), ...\}$.

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

From the relation R = $\{(x, y) : y = 2x, x, y \in N\}$, the first component is $x$. The condition $x \in N$ is explicitly given, and for every $x \in N$, $y = 2x$ is also in N. Thus, every natural number is the first component of some ordered pair in R.

Domain of R = $\{x : (x, y) \in R \text{ for some } y\}$

Domain of R = $\{x : x \in N\} = N$.

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The codomain of a relation from set A to set B is the set B. Here, the relation is defined on N, meaning from N to N. So, the codomain is N.

Codomain of R = N.

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

The second component is $y = 2x$, where $x \in \{1, 2, 3, ...\}$.

The values of $y$ are $\{2(1), 2(2), 2(3), ...\} = \{2, 4, 6, ...\}$.

Range of R = $\{y : (x, y) \in R \text{ for some } x\}$

Range of R = $\{y : y = 2x \text{ for some } x \in N\} = \{2, 4, 6, ...\}$ (the set of all even natural numbers).

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Is this relation a function?

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

In this case, the relation is on N, meaning it is a relation from N to N. So, set A = N and set B = N.

We need to check if every element $x \in N$ has exactly one image $y \in N$ such that $(x, y) \in R$.

The relation is defined by $y = 2x$. For every natural number $x$, the value $2x$ is a unique natural number $y$.

• Every element $x \in N$ is the first component of an ordered pair $(x, 2x)$ in R because $2x \in N$ for all $x \in N$.
• For a given $x \in N$, the value of $y = 2x$ is uniquely determined. There is only one $y$ associated with each $x$.

Thus, every element in the domain (which is N) has one and only one image in the codomain (which is N).

Therefore, the relation R is a function from N to N.

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Result:

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

Codomain of R = $\{1, 2, 3, ...\} = N$.

Range of R = $\{2, 4, 6, 8, ...\} = \{\text{all even natural numbers}\}$.

Yes, this relation is a function.

Given:

Three relations are given in roster form.

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

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

(iii) R = $\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)\}$.

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To Determine:

For each relation, state whether it is a function or not, and provide the reason.

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Solution:

A relation R from a set A to a set B is a function if and only if:

• Every element in the domain of the relation is from set A.
• Every element in the domain of the relation is associated with exactly one element in the codomain (set B). In other words, no two distinct ordered pairs in the relation have the same first component.

For relations given in roster form, we check if any element in the domain appears as the first component in more than one ordered pair.

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(i) R = {(2, 1), (3, 1), (4, 2)}

The domain of R is the set of first components: $\{2, 3, 4\}$.

Let's examine the first components:

• The element 2 is associated only with the image 1 (in the pair (2, 1)).
• The element 3 is associated only with the image 1 (in the pair (3, 1)).
• The element 4 is associated only with the image 2 (in the pair (4, 2)).

Each element in the domain $\{2, 3, 4\}$ is associated with exactly one image.

Therefore, the relation R is a function.

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(ii) R = {(2, 2), (2, 4), (3, 3), (4, 4)}

The domain of R is the set of first components: $\{2, 3, 4\}$.

Let's examine the first components:

• The element 2 appears as the first component in two ordered pairs: (2, 2) and (2, 4).

This means that the element 2 in the domain is associated with two distinct images (2 and 4) in the codomain.

According to the definition of a function, each element in the domain must have only one image.

Therefore, the relation R is not a function.

Reason: The element 2 in the domain has two images (2 and 4).

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(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}

The domain of R is the set of first components: $\{1, 2, 3, 4, 5, 6\}$.

Let's examine the first components:

• The element 1 is associated only with the image 2 (in the pair (1, 2)).
• The element 2 is associated only with the image 3 (in the pair (2, 3)).
• The element 3 is associated only with the image 4 (in the pair (3, 4)).
• The element 4 is associated only with the image 5 (in the pair (4, 5)).
• The element 5 is associated only with the image 6 (in the pair (5, 6)).
• The element 6 is associated only with the image 7 (in the pair (6, 7)).

Each element in the domain $\{1, 2, 3, 4, 5, 6\}$ is associated with exactly one image.

Therefore, the relation R is a function.

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Summary of Results:

(i) R = $\{(2, 1), (3, 1), (4, 2)\}$ is a function. Reason: Each element in the domain $\{2, 3, 4\}$ has a unique image.

(ii) R = $\{(2, 2), (2, 4), (3, 3), (4, 4)\}$ is not a function. Reason: The element 2 in the domain has two images (2 and 4).

(iii) R = $\{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)\}$ is a function. Reason: Each element in the domain $\{1, 2, 3, 4, 5, 6\}$ has a unique image.

Given:

Set of natural numbers N = $\{1, 2, 3, ...\}$

Real valued function f : N $\to$ N defined by $f(x) = 2x + 1$.

The table provides specific values for x from the domain of the function.

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To Complete:

The given table by finding the values of $y = f(x)$ for each given x.

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Solution:

The function rule is $f(x) = 2x + 1$. We need to substitute each given value of x into the function to find the corresponding value of $f(x)$.

• For $x = 1$: $f(1) = 2(1) + 1 = 2 + 1 = 3$
• For $x = 2$: $f(2) = 2(2) + 1 = 4 + 1 = 5$
• For $x = 3$: $f(3) = 2(3) + 1 = 6 + 1 = 7$
• For $x = 4$: $f(4) = 2(4) + 1 = 8 + 1 = 9$
• For $x = 5$: $f(5) = 2(5) + 1 = 10 + 1 = 11$
• For $x = 6$: $f(6) = 2(6) + 1 = 12 + 1 = 13$
• For $x = 7$: $f(7) = 2(7) + 1 = 14 + 1 = 15$

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Now, we can fill in the values in the table.

x	1	2	3	4	5	6	7
y = f(x)	3	5	7	9	11	13	15

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Result:

The completed table is shown above.

Given:

The function f : R $\to$ R is defined by $f(x) = x^2$, where $x \in R$.

A table with specific values of x is provided.

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To Find:

Complete the given table.

Find the domain and range of the function f.

Draw the graph of f.

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Solution:

The function is $f(x) = x^2$. We calculate the value of $y = f(x)$ for each given value of x in the table.

• For $x = -4$: $f(-4) = (-4)^2 = 16$
• For $x = -3$: $f(-3) = (-3)^2 = 9$
• For $x = -2$: $f(-2) = (-2)^2 = 4$
• For $x = -1$: $f(-1) = (-1)^2 = 1$
• For $x = 0$: $f(0) = (0)^2 = 0$
• For $x = 1$: $f(1) = (1)^2 = 1$
• For $x = 2$: $f(2) = (2)^2 = 4$
• For $x = 3$: $f(3) = (3)^2 = 9$
• For $x = 4$: $f(4) = (4)^2 = 16$

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The completed table is as follows:

x	–4	–3	–2	–1	0	1	2	3	4
y = f(x) = $x^2$	16	9	4	1	0	1	4	9	16

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Domain of f:

The domain of the function f is the set of all real numbers for which the function is defined. The given definition states that $x \in R$. For any real number x, $x^2$ is a well-defined real number.

Therefore, the domain of f is the set of all real numbers, R.

Domain(f) = R

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Range of f:

The range of the function f is the set of all possible values of $f(x)$ as x varies over the domain. For $f(x) = x^2$, the square of any real number is always greater than or equal to zero.

For any $x \in R$, $x^2 \ge 0$.

Also, for any non-negative real number $y \ge 0$, there exists a real number $x = \sqrt{y}$ such that $f(x) = (\sqrt{y})^2 = y$.

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

Range(f) = $\{y \in R : y \ge 0\}$ or $[0, \infty)$

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Graph of f:

The graph of the function $f(x) = x^2$ is a parabola. It opens upwards and is symmetric about the y-axis. The vertex of the parabola is at the origin $(0, 0)$.

Using the points calculated in the table (e.g., $(-4, 16), (-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9), (4, 16)$), we can plot these points on a Cartesian plane and connect them with a smooth curve to obtain the graph of the parabola $y = x^2$.

[Note: Drawing the actual graph image is not supported in this format. The description above explains the shape and key features of the graph.]

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Result:

The completed table is provided above.

Domain of f = R

Range of f = $\{y \in R : y \ge 0\}$ or $[0, \infty)$

The graph of f is an upward-opening parabola with its vertex at (0, 0).

Given:

Function f : R $\to$ R defined by $f(x) = x^3$, where $x \in R$.

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To Draw:

The graph of the function f.

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Solution:

The function is $f(x) = x^3$. To draw the graph, we can find some points $(x, f(x))$ that lie on the graph by choosing various values for x and calculating the corresponding $y = f(x)$ values.

Let's calculate $f(x)$ for a few values of x:

x	f(x) = $x^3$	Point (x, f(x))
-2	$(-2)^3 = -8$	(-2, -8)
-1	$(-1)^3 = -1$	(-1, -1)
0	$(0)^3 = 0$	(0, 0)
1	$(1)^3 = 1$	(1, 1)
2	$(2)^3 = 8$	(2, 8)

Other points could include: $(-3, -27), (3, 27)$, etc.

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The graph of $f(x) = x^3$ is a characteristic cubic curve.

It passes through the origin $(0, 0)$.

In the first quadrant ($x > 0$), the value of $f(x)$ is positive and increases rapidly as x increases.

In the third quadrant ($x < 0$), the value of $f(x)$ is negative and decreases rapidly (becomes more negative) as x decreases.

The graph is symmetric with respect to the origin because $f(-x) = (-x)^3 = -x^3 = -f(x)$, which is the property of an odd function.

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Domain and Range:

The domain of the function is given as R, the set of all real numbers. For any real number x, $x^3$ is a well-defined real number.

Domain(f) = R

The range of the function is the set of all possible values of $f(x)$. For any real number y, there exists a real number x (specifically, $x = \sqrt[3]{y}$) such that $f(x) = x^3 = y$.

Therefore, the range of the function is also the set of all real numbers.

Range(f) = R

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Instructions for Drawing the Graph:

1. Draw a Cartesian coordinate system with perpendicular x-axis and y-axis intersecting at the origin (0, 0).

2. Plot the points calculated above, such as $(-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)$. You may need to adjust the scale on the y-axis to accommodate larger values like 8 or 27.

3. Draw a smooth continuous curve that passes through these plotted points. The curve should extend infinitely upwards as x goes to positive infinity and infinitely downwards as x goes to negative infinity.

[Note: An actual image of the graph cannot be provided in this format. The description above helps in manually drawing or visualizing the graph.]

Given:

The real valued function f : R – $\{0\}$ $\to$ R defined by $f(x) = \frac{1}{x}$, where $x \in R - \{0\}$.

A table with specific values of x from the domain is provided.

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To Find:

Complete the given table.

Find the domain and range of the function f.

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Solution:

The function rule is $f(x) = \frac{1}{x}$. We need to calculate the value of $y = f(x)$ for each given value of x in the table.

• For $x = -2$: $f(-2) = \frac{1}{-2} = -0.5$
• For $x = -1.5$: $f(-1.5) = \frac{1}{-1.5} = \frac{1}{-3/2} = -\frac{2}{3}$
• For $x = -1$: $f(-1) = \frac{1}{-1} = -1$
• For $x = -0.5$: $f(-0.5) = \frac{1}{-0.5} = \frac{1}{-1/2} = -2$
• For $x = 0.25$: $f(0.25) = \frac{1}{0.25} = \frac{1}{1/4} = 4$
• For $x = 0.5$: $f(0.5) = \frac{1}{0.5} = \frac{1}{1/2} = 2$
• For $x = 1$: $f(1) = \frac{1}{1} = 1$
• For $x = 1.5$: $f(1.5) = \frac{1}{1.5} = \frac{1}{3/2} = \frac{2}{3}$
• For $x = 2$: $f(2) = \frac{1}{2} = 0.5$

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The completed table is as follows:

x	–2	–1.5	–1	–0.5	0.25	0.5	1	1.5	2
$y = \frac{1}{x}$	-0.5	$-\frac{2}{3}$	-1	-2	4	2	1	$\frac{2}{3}$	0.5

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Domain of f:

The domain of the function is the set of all input values x for which the function is defined. The given definition explicitly states that the function is defined for $x \in R - \{0\}$. This is because division by zero is undefined.

Therefore, the domain of f is the set of all real numbers except 0.

Domain(f) = R - $\{0\}$

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Range of f:

The range of the function is the set of all possible output values $f(x)$ as x varies over the domain. The function is $f(x) = \frac{1}{x}$ where $x \in R - \{0\}$.

Consider any non-zero real number $y$. Can we find an $x \in R - \{0\}$ such that $f(x) = y$?

We need to solve $y = \frac{1}{x}$ for x. Multiplying both sides by x (which is non-zero), we get $yx = 1$. Since $y \neq 0$, we can divide by y to get $x = \frac{1}{y}$.

Since $y$ is a non-zero real number, $\frac{1}{y}$ is also a non-zero real number. So, for any non-zero real number y, there exists a corresponding non-zero real number x such that $f(x) = y$.

The only value that $f(x) = \frac{1}{x}$ can never be is 0, because $\frac{1}{x} = 0$ has no solution for any finite x.

Therefore, the range of f is the set of all real numbers except 0.

Range(f) = R - $\{0\}$

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Result:

The completed table is provided above.

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

Range of f = R - $\{0\}$

[Note: Drawing the graph of this function (a hyperbola) involves plotting points from the table and understanding the asymptotic behavior at x=0 and y=0. A visual representation is not possible in this format.]

Given:

Two real functions f and g defined as:

$f(x) = x^2$

$g(x) = 2x + 1$

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To Find:

$(f + g)(x)$

$(f - g)(x)$

$(fg)(x)$

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

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Solution:

For two real functions f and g, the operations are defined as follows:

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

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

$(fg)(x) = f(x)g(x)$

$\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)}$, provided $g(x) \neq 0$.

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(f + g) (x):

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

Substitute the given definitions of f(x) and g(x):

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

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

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(f – g) (x):

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

Substitute the given definitions of f(x) and g(x):

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

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

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(fg) (x):

$(fg)(x) = f(x)g(x)$

Substitute the given definitions of f(x) and g(x):

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

Distribute $x^2$ into the parentheses:

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

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

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

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

Substitute the given definitions of f(x) and g(x):

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

This expression is defined for all real numbers x except when the denominator is zero.

The denominator is zero when $2x + 1 = 0$, which means $2x = -1$, so $x = -\frac{1}{2}$.

Thus, the domain of $\left( \frac{f}{g} \right)(x)$ is R - $\{-\frac{1}{2}\}$.

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Result:

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

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

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

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

Given:

Two functions f and g defined over the set of non-negative real numbers.

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

$g(x) = x$

The domain for both functions is the set of non-negative real numbers, denoted as $[0, \infty)$.

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To Find:

$(f + g)(x)$

$(f - g)(x)$

$(fg)(x)$

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

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Solution:

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

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

This is defined for all x in the intersection of the domains of f and g. Since both are defined for $x \ge 0$, the domain of $(f+g)(x)$ is $[0, \infty)$.

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

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

This is defined for all x in the intersection of the domains of f and g, which is $[0, \infty)$.

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

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

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

$(fg)(x) = x^{1/2 + 1}$

$(fg)(x) = x^{3/2}$.

This is defined for all x in the intersection of the domains of f and g, which is $[0, \infty)$.

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

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

This is defined for all x where both $f(x)$ and $g(x)$ are defined, AND $g(x) \neq 0$.

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

Domain of $g(x)$ is $[0, \infty)$ (as per the problem statement, defined over non-negative reals).

The condition $g(x) \neq 0$ means $x \neq 0$.

So, the domain of $\left( \frac{f}{g} \right)(x)$ is $[0, \infty)$ excluding $\{0\}$, which is $(0, \infty)$.

For $x > 0$, we can simplify the expression:

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

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Result:

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

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

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

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

To Determine:

Which of the given relations are functions. Provide reasons and find the domain and range if it is a function.

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Solution:

A relation is a function if and only if every element in its domain has one and only one image. In terms of ordered pairs, this means that no two distinct ordered pairs in the relation have the same first component.

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(i) R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

The first components of the ordered pairs are 2, 5, 8, 11, 14, and 17.

Each of these first components appears exactly once in the relation.

Therefore, every element in the domain is associated with a unique image.

Conclusion: The relation is a function.

Domain of R = Set of all first components = $\{2, 5, 8, 11, 14, 17\}$.

Range of R = Set of all second components = $\{1\}$.

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(ii) R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

The first components of the ordered pairs are 2, 4, 6, 8, 10, 12, and 14.

Each of these first components appears exactly once in the relation.

Therefore, every element in the domain is associated with a unique image.

Conclusion: The relation is a function.

Domain of R = Set of all first components = $\{2, 4, 6, 8, 10, 12, 14\}$.

Range of R = Set of all second components = $\{1, 2, 3, 4, 5, 6, 7\}$.

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(iii) R = {(1, 3), (1, 5), (2, 5)}

The first components of the ordered pairs are 1, 1, and 2.

The first component 1 appears in two different ordered pairs: (1, 3) and (1, 5).

This means that the element 1 is associated with two different images (3 and 5).

According to the definition of a function, each element in the domain must have only one image.

Conclusion: The relation is not a function.

Reason: The element 1 in the domain is related to two distinct images (3 and 5).

(i) $f(x) = -|x|$

Domain:

The function $f(x) = -|x|$ is defined for all real values of $x$ because the absolute value function $|x|$ is defined for all real numbers, and multiplying by -1 does not change the domain.

So, the domain is the set of all real numbers, denoted by $\mathbb{R}$ or $(-\infty, \infty)$.

Range:

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

Multiplying by -1 reverses the inequality, so $-|x| \le 0$.

Thus, the value of $f(x)$ can be any non-positive real number.

So, the range is the set of all non-positive real numbers, denoted by $(-\infty, 0]$.

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(ii) $f(x) = \sqrt{9 - x^2}$

Domain:

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

So, we must have $9 - x^2 \ge 0$.

This inequality can be rewritten as $x^2 \le 9$.

Taking the square root of both sides (and considering both positive and negative roots), we get $-\sqrt{9} \le x \le \sqrt{9}$, which means $-3 \le x \le 3$.

So, the domain is the closed interval $[-3, 3]$.

Range:

For the domain $[-3, 3]$, the value of $x^2$ ranges from $0$ (when $x=0$) to $9$ (when $x=3$ or $x=-3$).

So, $0 \le x^2 \le 9$.

Multiplying by -1 reverses the inequality: $-9 \le -x^2 \le 0$.

Adding 9 to all parts of the inequality: $9 - 9 \le 9 - x^2 \le 9 - 0$, which simplifies to $0 \le 9 - x^2 \le 9$.

Taking the square root of all parts (since all parts are non-negative): $\sqrt{0} \le \sqrt{9 - x^2} \le \sqrt{9}$.

This gives $0 \le f(x) \le 3$.

So, the range is the closed interval $[0, 3]$.

The given function is $f(x) = 2x - 5$.

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(i) To find $f(0)$, substitute $x = 0$ into the function:

$f(0) = 2(0) - 5$

$f(0) = 0 - 5$

$f(0) = -5$

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(ii) To find $f(7)$, substitute $x = 7$ into the function:

$f(7) = 2(7) - 5$

$f(7) = 14 - 5$

$f(7) = 9$

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(iii) To find $f(-3)$, substitute $x = -3$ into the function:

$f(-3) = 2(-3) - 5$

$f(-3) = -6 - 5$

$f(-3) = -11$

The given function is $t(C) = \frac{9C}{5} + 32$, where C is the temperature in degree Celsius and t(C) is the temperature in degree Fahrenheit.

---

(i) To find $t(0)$, substitute $C = 0$ into the function:

$t(0) = \frac{9(0)}{5} + 32$

$t(0) = 0 + 32$

$t(0) = 32$

---

(ii) To find $t(28)$, substitute $C = 28$ into the function:

$t(28) = \frac{9(28)}{5} + 32$

$t(28) = \frac{252}{5} + 32$

$t(28) = 50.4 + 32$

$t(28) = 82.4$

---

(iii) To find $t(-10)$, substitute $C = -10$ into the function:

$t(-10) = \frac{9(-10)}{5} + 32$

$t(-10) = \frac{-90}{5} + 32$

$t(-10) = -18 + 32$

$t(-10) = 14$

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(iv) To find the value of C when $t(C) = 212$, set the function equal to 212 and solve for C:

$t(C) = 212$

$\frac{9C}{5} + 32 = 212$

Subtract 32 from both sides:

$\frac{9C}{5} = 212 - 32$

$\frac{9C}{5} = 180$

Multiply both sides by 5:

$9C = 180 \times 5$

$9C = 900$

Divide both sides by 9:

$C = \frac{900}{9}$

$C = 100$

So, when the temperature in Fahrenheit is 212$^\circ$F, the temperature in Celsius is 100$^\circ$C.

(i) $f(x) = 2 - 3x$, $x \in \mathbb{R}$, $x > 0$.

Given the domain is $x > 0$.

To find the range, we look at the possible values of $f(x)$.

Since $x > 0$, multiplying by -3 reverses the inequality:

$-3x < 0$

Now, add 2 to both sides:

$2 - 3x < 2 + 0$

$2 - 3x < 2$

So, $f(x) < 2$.

The range is the set of all real numbers less than 2.

Range: $(-\infty, 2)$ or $\{y \in \mathbb{R} \mid y < 2\}$.

---

(ii) $f(x) = x^2 + 2$, $x$ is a real number.

Given the domain is all real numbers, i.e., $x \in \mathbb{R}$.

To find the range, we consider the possible values of $f(x)$.

For any real number $x$, we know that $x^2 \ge 0$.

Adding 2 to both sides:

$x^2 + 2 \ge 0 + 2$

$x^2 + 2 \ge 2$

So, $f(x) \ge 2$.

The range is the set of all real numbers greater than or equal to 2.

Range: $[2, \infty)$ or $\{y \in \mathbb{R} \mid y \ge 2\}$.

---

(iii) $f(x) = x$, $x$ is a real number.

Given the domain is all real numbers, i.e., $x \in \mathbb{R}$.

This is the identity function. For every real number input $x$, the output $f(x)$ is exactly the same real number $x$.

Therefore, the set of all possible output values is the set of all real numbers.

Range: $\mathbb{R}$ or $(-\infty, \infty)$.

The given function is $f(x) = x + 10$.

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

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

---

Since the function $f(x) = x + 10$ is a linear function of the form $y = mx + c$ (where $m=1$ and $c=10$), its graph will be a straight line.

To sketch the graph of a linear function, we need to find at least two points that lie on the line. We can choose any values for $x$ from the domain and find the corresponding values of $f(x)$.

---

Let's choose a few values for $x$ and calculate $f(x)$:

x	f(x) = x + 10	Point (x, f(x))
-2	-2 + 10 = 8	(-2, 8)
0	0 + 10 = 10	(0, 10)
2	2 + 10 = 12	(2, 12)
5	5 + 10 = 15	(5, 15)

---

To sketch the graph:

1. Draw a coordinate plane with the x-axis and the y-axis.

2. Plot the points found in the table, such as (-2, 8), (0, 10), (2, 12), and (5, 15).

3. Draw a straight line passing through these points.

This straight line represents the graph of the function $f(x) = x + 10$. The line will have a slope of 1 and a y-intercept of 10.

Given:

R is a relation from the set of rational numbers $\mathbb{Q}$ to $\mathbb{Q}$, defined by R = {(a,b): a,b $\in$ $\mathbb{Q}$ and a – b $\in$ $\mathbb{Z}$}.

---

To Show:

(i) (a, a) $\in$ R for all a $\in$ $\mathbb{Q}$.

(ii) (a, b) $\in$ R implies that (b, a) $\in$ R.

(iii) (a, b) $\in$ R and (b,c) $\in$ R implies that (a, c) $\in$ R.

---

Proof:

(i) To show (a, a) $\in$ R for all a $\in$ $\mathbb{Q}$:

According to the definition of the relation R, $(a, a) \in R$ if and only if the difference $a - a$ is an integer, i.e., $a - a \in \mathbb{Z}$.

For any rational number $a \in \mathbb{Q}$, the difference $a - a$ is always 0.

Since 0 is an integer, $0 \in \mathbb{Z}$.

Therefore, $a - a \in \mathbb{Z}$ for all $a \in \mathbb{Q}$.

Hence, by the definition of R, (a, a) $\in$ R for all a $\in$ $\mathbb{Q}$.

---

(ii) To show that (a, b) $\in$ R implies that (b, a) $\in$ R:

Assume that $(a, b) \in R$.

By the definition of the relation R, $(a, b) \in R$ means that the difference $a - b$ is an integer.

We want to show that $(b, a) \in R$. By the definition of R, this means we need to show that $b - a \in \mathbb{Z}$.

Consider the expression $b - a$. This is the negative of $a - b$.

$b - a = -(a - b)$

Since $a - b$ is an integer (from our assumption), its negative, $-(a - b)$, must also be an integer.

Thus, $b - a \in \mathbb{Z}$.

Therefore, by the definition of R, (b, a) $\in$ R.

Hence, (a, b) $\in$ R implies that (b, a) $\in$ R.

---

(iii) To show that (a, b) $\in$ R and (b,c) $\in$ R implies that (a, c) $\in$R:

Assume that $(a, b) \in R$ and $(b, c) \in R$.

From $(a, b) \in R$, by the definition of R, we have $a - b \in \mathbb{Z}$. Let $a - b = k_1$, where $k_1$ is some integer ($k_1 \in \mathbb{Z}$).

From $(b, c) \in R$, by the definition of R, we have $b - c \in \mathbb{Z}$. Let $b - c = k_2$, where $k_2$ is some integer ($k_2 \in \mathbb{Z}$).

We want to show that $(a, c) \in R$. By the definition of R, this means we need to show that $a - c \in \mathbb{Z}$.

Consider the sum of the two differences we have:

Simplifying the left side, we get $a - c$.

Since $k_1$ and $k_2$ are integers, their sum $k_1 + k_2$ is also an integer.

Thus, $a - c \in \mathbb{Z}$.

Therefore, by the definition of R, (a, c) $\in$ R.

Hence, (a, b) $\in$ R and (b,c) $\in$ R implies that (a, c) $\in$R.

Given that $f$ is a linear function from $\mathbb{Z}$ into $\mathbb{Z}$. A linear function can be written in the form $f(x) = ax + b$, where $a$ and $b$ are constants (in this case, integers since the codomain is $\mathbb{Z}$).

We are given a set of points that lie on the graph of this function:

$f = \{(1, 1), (2, 3), (0, –1), (–1, –3)\}$

Each pair $(x, f(x))$ satisfies the equation $f(x) = ax + b$. We can use any two distinct points to set up a system of linear equations to find the values of $a$ and $b$.

---

Let's use the point (0, -1). This means when $x = 0$, $f(x) = -1$.

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

$f(0) = a(0) + b$

From equation (i), we get $b = -1$.

---

Now let's use another point, say (1, 1). This means when $x = 1$, $f(x) = 1$.

Substitute these values and the value of $b = -1$ into $f(x) = ax + b$:

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

From equation (ii), we get:

1 = a - 1

a = 1 + 1

a = 2

---

So, the values of the constants are $a = 2$ and $b = -1$.

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

Thus, the function is $f(x) = 2x - 1$.

---

We can verify this function with the other given points:

For the point (2, 3): $f(2) = 2(2) - 1 = 4 - 1 = 3$. This matches the given point.

For the point (-1, -3): $f(-1) = 2(-1) - 1 = -2 - 1 = -3$. This matches the given point.

Since the function $f(x) = 2x - 1$ holds true for all given points, this is the correct linear function.

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

The given function is $f(x) = \frac{x^{2} + 3x + 5}{x^{2} - 5x + 4}$.

---

For a rational function (a function expressed as a ratio of two polynomials) to be defined, the denominator cannot be equal to zero.

Therefore, to find the domain of $f(x)$, we need to find the values of $x$ for which the denominator $x^2 - 5x + 4$ is zero and exclude those values from the set of all real numbers.

---

Set the denominator equal to zero:

$x^2 - 5x + 4 = 0$

This is a quadratic equation. We can solve it by factoring.

We need two numbers that multiply to 4 and add up to -5. These numbers are -1 and -4.

So, we can factor the quadratic expression as:

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

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

Case 1:

$x - 1 = 0$

$x = 1$

Case 2:

$x - 4 = 0$

$x = 4$

---

Thus, the denominator is zero when $x = 1$ or $x = 4$. These values must be excluded from the domain of the function.

The domain of $f(x)$ is the set of all real numbers except 1 and 4.

In set notation, the domain is $\{x \in \mathbb{R} \mid x \neq 1 \text{ and } x \neq 4\}$.

In interval notation, the domain is $(-\infty, 1) \cup (1, 4) \cup (4, \infty)$.

The domain of the function is $\mathbb{R} \setminus \{1, 4\}$.

The given function is a piecewise function defined as:

$f(x) = \begin{cases} 1 - x & , & x < 0 \\ 1 & , & x = 0\\ x + 1 & , & x > 0 \end{cases}$

To draw the graph of this function, we consider each part of the definition based on the specified domain for $x$.

---

Case 1: For $x < 0$, $f(x) = 1 - x$.

This part of the function is a linear equation, $y = 1 - x$, for $x$ values less than 0. The graph is a straight line with a slope of -1 and a y-intercept of 1 (though the y-intercept itself is not included in this specific domain). To plot this part, we can take a few values of $x < 0$ and find the corresponding $f(x)$ values.

If $x = -1$, $f(-1) = 1 - (-1) = 1 + 1 = 2$. Point: $(-1, 2)$.

If $x = -2$, $f(-2) = 1 - (-2) = 1 + 2 = 3$. Point: $(-2, 3)$.

As $x$ approaches 0 from the left ($x \to 0^-$), $f(x) = 1 - x$ approaches $1 - 0 = 1$. So, the line segment for $x < 0$ extends up to the point $(0, 1)$, but this point is not included in this part's domain, so we use an open circle at $(0, 1)$.

---

Case 2: For $x = 0$, $f(x) = 1$.

This part of the function defines a single point on the graph. When $x$ is exactly 0, the value of the function is 1. This gives the point $(0, 1)$. This point is included in the graph, so we plot it as a closed circle.

---

Case 3: For $x > 0$, $f(x) = x + 1$.

This part of the function is also a linear equation, $y = x + 1$, for $x$ values greater than 0. The graph is a straight line with a slope of 1 and a y-intercept of 1 (though the y-intercept itself is not included in this specific domain). To plot this part, we can take a few values of $x > 0$ and find the corresponding $f(x)$ values.

If $x = 1$, $f(1) = 1 + 1 = 2$. Point: $(1, 2)$.

If $x = 2$, $f(2) = 2 + 1 = 3$. Point: $(2, 3)$.

As $x$ approaches 0 from the right ($x \to 0^+$), $f(x) = x + 1$ approaches $0 + 1 = 1$. So, the line segment for $x > 0$ starts from the point $(0, 1)$, but this point is not included in this part's domain, so we use an open circle at $(0, 1)$.

---

Combining the parts:

The graph of the function consists of three parts:

1. A ray extending from $(0, 1)$ (exclusive) to the left and upwards, corresponding to $y = 1 - x$ for $x < 0$.

2. A single point at $(0, 1)$ (inclusive).

3. A ray extending from $(0, 1)$ (exclusive) to the right and upwards, corresponding to $y = x + 1$ for $x > 0$.

Notice that the open circles from Case 1 and Case 3 are both at $(0, 1)$. The closed circle from Case 2 is also at $(0, 1)$. This means the three parts connect seamlessly at the point $(0, 1)$.

The graph will be a "V" shape with its vertex (or point) at $(0, 1)$. The left arm of the "V" ($x < 0$) has a slope of -1, and the right arm ($x > 0$) has a slope of 1.

To draw the graph, plot the point $(0, 1)$. Then draw a straight line passing through $(0, 1)$ and $(-1, 2)$ and $(-2, 3)$, extending indefinitely to the left. Also, draw a straight line passing through $(0, 1)$ and $(1, 2)$ and $(2, 3)$, extending indefinitely to the right.

A relation is a function if and only if for every element in its domain, there is exactly one corresponding element in its range.

---

Let's analyze the relation $f$ defined by:

$f(x) = \begin{cases} x^2 & , & 0 \leq x \leq 3 \\ 3x & , & 3 \leq x \leq 10 \end{cases}$

The domain of $f$ is specified as the union of the intervals $[0, 3]$ and $[3, 10]$, which is $[0, 10]$.

For any $x$ in the interval $[0, 3)$ (i.e., $0 \leq x < 3$), the value of $f(x)$ is uniquely given by $f(x) = x^2$.

For any $x$ in the interval $(3, 10]$ (i.e., $3 < x \leq 10$), the value of $f(x)$ is uniquely given by $f(x) = 3x$.

We need to check the point where the definitions meet, which is $x = 3$.

According to the first part of the definition, if $x=3$ (since $0 \leq 3 \leq 3$), $f(3) = 3^2 = 9$.

According to the second part of the definition, if $x=3$ (since $3 \leq 3 \leq 10$), $f(3) = 3(3) = 9$.

At $x=3$, both parts of the definition give the same value, $f(3)=9$.

Since for every value of $x$ in the domain $[0, 10]$, there is exactly one corresponding value of $f(x)$, the relation $f$ is a function.

---

Let's analyze the relation $g$ defined by:

$g(x) = \begin{cases} x^2 & , & 0 \leq x \leq 2 \\ 3x & , & 2 \leq x \leq 10 \end{cases}$

The domain of $g$ is specified as the union of the intervals $[0, 2]$ and $[2, 10]$, which is $[0, 10]$.

For any $x$ in the interval $[0, 2)$ (i.e., $0 \leq x < 2$), the value of $g(x)$ is uniquely given by $g(x) = x^2$.

For any $x$ in the interval $(2, 10]$ (i.e., $2 < x \leq 10$), the value of $g(x)$ is uniquely given by $g(x) = 3x$.

We need to check the point where the definitions meet, which is $x = 2$.

According to the first part of the definition, if $x=2$ (since $0 \leq 2 \leq 2$), $g(2) = 2^2 = 4$.

According to the second part of the definition, if $x=2$ (since $2 \leq 2 \leq 10$), $g(2) = 3(2) = 6.

At $x=2$, the two parts of the definition give different values ($4$ and $6$). This means that for the input $x=2$, the relation $g$ assigns two different outputs, which violates the condition for a function.

Since there exists an element in the domain (specifically $x=2$) that is mapped to more than one element in the range, the relation $g$ is not a function.

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

We need to find the value of $\frac{f(1.1) - f(1)}{(1.1 - 1)}$.

---

First, let's find the value of $f(1.1)$ by substituting $x = 1.1$ into the function:

$f(1.1) = (1.1)^2$

$f(1.1) = 1.21$

---

Next, let's find the value of $f(1)$ by substituting $x = 1$ into the function:

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

$f(1) = 1$

---

Now, let's calculate the numerator $f(1.1) - f(1)$:

$f(1.1) - f(1) = 1.21 - 1$

$f(1.1) - f(1) = 0.21$

---

Next, let's calculate the denominator $1.1 - 1$:

$1.1 - 1 = 0.1$

---

Finally, divide the numerator by the denominator:

$\frac{f(1.1) - f(1)}{(1.1 - 1)} = \frac{0.21}{0.1}$

To divide by 0.1, we can multiply both the numerator and the denominator by 10:

$\frac{0.21 \times 10}{0.1 \times 10} = \frac{2.1}{1}$

$\frac{0.21}{0.1} = 2.1$

---

Thus, the value of $\frac{f(1.1) - f(1)}{(1.1 - 1)}$ is 2.1.

The given function is $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

---

For a real function defined as a fraction, the domain is the set of all real numbers for which the denominator is not equal to zero.

Therefore, to find the domain of $f(x)$, we must exclude the values of $x$ that make the denominator $x^2 - 8x + 12$ equal to zero.

---

Set the denominator to zero:

$x^2 - 8x + 12 = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 12 and add up to -8. These numbers are -2 and -6.

So, we can factor the quadratic expression as:

$(x - 2)(x - 6) = 0$

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

Set each factor equal to zero:

$x - 2 = 0 \quad \text{or} \quad x - 6 = 0$

Solving for $x$ in each case:

$x = 2 \quad \text{or} \quad x = 6$

---

These are the values of $x$ for which the denominator is zero, and thus the function is undefined.

The domain of the function $f(x)$ is the set of all real numbers except 2 and 6.

In set notation, the domain is $\{x \in \mathbb{R} \mid x \neq 2 \text{ and } x \neq 6\}$.

In interval notation, the domain is $(-\infty, 2) \cup (2, 6) \cup (6, \infty)$.

The domain of the function is $\mathbb{R} \setminus \{2, 6\}$.

The given function is $f(x) = \sqrt{x - 1}$.

---

Domain:

For the function $f(x)$ to be a real function, the expression under the square root must be non-negative.

So, we must have:

$x - 1 \ge 0$

Adding 1 to both sides of the inequality:

$x \ge 1$

Thus, the domain of the function is the set of all real numbers greater than or equal to 1.

In set notation, the domain is $\{x \in \mathbb{R} \mid x \ge 1\}$.

In interval notation, the domain is $[1, \infty)$.

The domain of the function is $[1, \infty)$.

---

Range:

For the domain $x \ge 1$, we have $x - 1 \ge 0$.

The square root of a non-negative number is always non-negative.

So, $\sqrt{x - 1} \ge \sqrt{0}$

$\sqrt{x - 1} \ge 0$

Since $f(x) = \sqrt{x - 1}$, we have $f(x) \ge 0$.

Also, as $x$ increases, $\sqrt{x-1}$ increases. The values of $\sqrt{x-1}$ can be arbitrarily large as $x$ becomes very large.

Thus, the range of the function is the set of all non-negative real numbers.

In set notation, the range is $\{y \in \mathbb{R} \mid y \ge 0\}$.

In interval notation, the range is $[0, \infty)$.

The range of the function is $[0, \infty)$.

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

---

Domain:

The function $f(x) = |x - 1|$ involves the absolute value of the expression $(x - 1)$. The expression $(x - 1)$ is defined for all real numbers $x$. The absolute value function $|y|$ is defined for all real numbers $y$.

Therefore, the function $f(x) = |x - 1|$ is defined for all real values of $x$.

So, the domain is the set of all real numbers, denoted by $\mathbb{R}$ or $(-\infty, \infty)$.

The domain of the function is $\mathbb{R}$.

---

Range:

The absolute value of any real number is always non-negative. That is, for any real number $y$, $|y| \ge 0$.

In this function, the expression inside the absolute value is $x - 1$. Regardless of the value of $x$, the value of $x - 1$ is a real number.

Therefore, $|x - 1| \ge 0$ for all real numbers $x$.

The value of $f(x)$ can be any non-negative real number. For instance, if we want $f(x) = 5$, we can have $|x-1| = 5$, which means $x-1=5$ (so $x=6$) or $x-1=-5$ (so $x=-4$). Both 6 and -4 are in the domain, and $f(6)=|6-1|=5$, $f(-4)=|-4-1|=|-5|=5$.

So, the range is the set of all non-negative real numbers, denoted by $[0, \infty)$.

The range of the function is $[0, \infty)$.

The given function is $f(x) = \frac{x^2}{1 + x^2}$. The domain of the function is the set of all real numbers, $\mathbb{R}$.

To determine the range of $f$, we need to find the set of all possible values that $f(x)$ can take for $x \in \mathbb{R}$.

---

Let $y = f(x) = \frac{x^2}{1 + x^2}$. We want to find the possible values of $y$.

Consider the numerator $x^2$ and the denominator $1 + x^2$.

For any real number $x$, $x^2 \ge 0$.

Also, for any real number $x$, $x^2 \ge 0$, which implies $1 + x^2 \ge 1$.

Since $1 + x^2 \ge 1$, the denominator is always positive.

Since $x^2 \ge 0$ and $1 + x^2 > 0$, the fraction $\frac{x^2}{1 + x^2}$ must be non-negative.

$f(x) = \frac{x^2}{1 + x^2} \ge 0$

The minimum value occurs when $x^2$ is minimum, which is 0 (at $x=0$).

$f(0) = \frac{0^2}{1 + 0^2} = \frac{0}{1} = 0$.

So, the lower bound for the range is 0, and this value is attained.

---

Now let's consider the upper bound. We can rewrite the function:

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

We can express the numerator in terms of the denominator:

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

Separate the terms:

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

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

We know that $x^2 \ge 0$ for all real $x$.

So, $1 + x^2 \ge 1$.

Taking the reciprocal of a number greater than or equal to 1 gives a value between 0 and 1 (inclusive of 1 only if $1+x^2=1$).

Since $1 + x^2 \ge 1$, we have $0 < \frac{1}{1 + x^2} \le 1$. Note that $\frac{1}{1+x^2}$ cannot be 0 because $1+x^2$ is never infinite.

Now, consider $-\frac{1}{1 + x^2}$. Multiplying the inequality $0 < \frac{1}{1 + x^2} \le 1$ by -1 reverses the direction of the inequalities:

$-1 \le - \frac{1}{1 + x^2} < 0$

Finally, add 1 to all parts of the inequality:

$1 - 1 \le 1 - \frac{1}{1 + x^2} < 1 + 0$

$0 \le f(x) < 1$

This shows that the values of $f(x)$ are always greater than or equal to 0 and strictly less than 1.

The value 0 is attained at $x=0$. The value 1 is never attained for any real $x$, because $x^2$ can never be equal to $1+x^2$.

Thus, the range of the function is the interval $[0, 1)$.

The range of f is $[0, 1)$.

The given functions are $f(x) = x + 1$ and $g(x) = 2x - 3$. Both functions are defined from $\mathbb{R}$ to $\mathbb{R}$, so their domain is $\mathbb{R}$.

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Sum of functions (f + g):

The sum of two functions $(f + g)(x)$ is defined as the sum of their values at $x$.

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

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

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

Combine like terms:

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

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

The domain of $(f + g)$ is the intersection of the domains of $f$ and $g$, which is $\mathbb{R} \cap \mathbb{R} = \mathbb{R}$.

So, $(f + g)(x) = 3x - 2$, with domain $\mathbb{R}$.

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Difference of functions (f – g):

The difference of two functions $(f - g)(x)$ is defined as the difference of their values at $x$.

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

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

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

Distribute the negative sign:

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

Combine like terms:

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

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

The domain of $(f - g)$ is the intersection of the domains of $f$ and $g$, which is $\mathbb{R} \cap \mathbb{R} = \mathbb{R}$.

So, $(f - g)(x) = -x + 4$, with domain $\mathbb{R}$.

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Quotient of functions ($\frac{f}{g}$):

The quotient of two functions $(\frac{f}{g})(x)$ is defined as the ratio of their values at $x$, provided the denominator is not zero.

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

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

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

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

The domains of $f$ and $g$ are both $\mathbb{R}$.

Find the values of $x$ where $g(x) = 0$:

$2x - 3 = 0$

$2x = 3$

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

So, the function $(\frac{f}{g})(x)$ is defined for all real numbers except $x = \frac{3}{2}$.

The domain of $(\frac{f}{g})$ is $\mathbb{R} \setminus \{\frac{3}{2}\}$.

So, $(\frac{f}{g})(x) = \frac{x + 1}{2x - 3}$, with domain $\mathbb{R} \setminus \{\frac{3}{2}\}$.

The given function is a linear function from $\mathbb{Z}$ to $\mathbb{Z}$, defined by $f(x) = ax + b$, where $a$ and $b$ are integers.

We are given a set of points that belong to this function: $f = \{(1, 1), (2, 3), (0, –1), (–1, –3)\}$.

Each ordered pair $(x, y)$ in the set means that $f(x) = y$. We can use any two points from this set to form a system of linear equations and solve for $a$ and $b$.

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Let's use the point (0, -1). This means when $x = 0$, $f(x) = -1$.

Substitute these values into the function definition $f(x) = ax + b$:

$f(0) = a(0) + b$

$-1 = 0 + b$

$b = -1$

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Now that we know $b = -1$, the function is $f(x) = ax - 1$.

Let's use another point, for example (1, 1). This means when $x = 1$, $f(x) = 1$.

Substitute these values and $b = -1$ into $f(x) = ax + b$:

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

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

$1 = a - 1$

Add 1 to both sides:

$1 + 1 = a$

$a = 2$

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So, the values of the integers $a$ and $b$ are $a = 2$ and $b = -1$.

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

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We can verify this by checking if the other given points satisfy this function:

For point (2, 3): $f(2) = 2(2) - 1 = 4 - 1 = 3$. This matches the given point (2, 3).

For point (-1, -3): $f(-1) = 2(-1) - 1 = -2 - 1 = -3$. This matches the given point (-1, -3).

Since the function $f(x) = 2x - 1$ is satisfied by all given points, our values for $a$ and $b$ are correct.

The determined values are $a = 2$ and $b = -1$.

The relation R is defined on the set of natural numbers $\mathbb{N}$ as $R = \{(a, b) : a, b \in \mathbb{N} \text{ and } a = b^2\}$. Let's check each statement.

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(i) Is (a, a) $\in$ R, for all a $\in$ $\mathbb{N}$?

For (a, a) to be in R, according to the definition, the first element must be the square of the second element. So, we must have $a = a^2$ for all $a \in \mathbb{N}$.

Let's test some natural numbers:

If $a = 1$, then $1 = 1^2$, which is true.

If $a = 2$, then $2 = 2^2$, which is $2 = 4$, false.

If $a = 3$, then $3 = 3^2$, which is $3 = 9$, false.

The condition $a = a^2$ is only true for $a=1$ (and $a=0$, but 0 is not always included in $\mathbb{N}$, and even if it were, it's not true for *all* $a \in \mathbb{N}$).

Since $a = a^2$ is not true for all $a \in \mathbb{N}$ (e.g., it is false for $a=2$), the statement is false.

Justification: Take $a = 2 \in \mathbb{N}$. We check if $(2, 2) \in R$. This would require $2 = 2^2$, which is $2 = 4$. This is false. Therefore, $(2, 2) \notin R$. Hence, $(a, a) \in R$ is not true for all $a \in \mathbb{N}$.

The statement (i) is False.

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(ii) If (a, b) $\in$ R, does it imply (b, a) $\in$ R?

Assume $(a, b) \in R$. By the definition of R, this means $a = b^2$ for $a, b \in \mathbb{N}$.

For $(b, a)$ to be in R, we would need $b = a^2$ for $a, b \in \mathbb{N}$.

Let's see if $a = b^2$ implies $b = a^2$ for $a, b \in \mathbb{N}$.

Consider an example: Let $b = 2 \in \mathbb{N}$. Then $a = b^2 = 2^2 = 4$. So, $a = 4 \in \mathbb{N}$.

We have $(4, 2) \in R$ because $4 = 2^2$ and $4, 2 \in \mathbb{N}$.

Now let's check if $(2, 4) \in R$. This would require $2 = 4^2$, which is $2 = 16$. This is false.

So, $(a, b) \in R$ does not necessarily imply $(b, a) \in R$.

Justification: Take $(4, 2)$. Here $a=4, b=2$. Since $4, 2 \in \mathbb{N}$ and $4 = 2^2$, we have $(4, 2) \in R$. Now check if $(2, 4) \in R$. This would require $2 = 4^2$, which is $2 = 16$. This is false. Therefore, $(2, 4) \notin R$. Hence, $(a, b) \in R$ does not imply $(b, a) \in R$.

The statement (ii) is False.

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(iii) If (a, b) $\in$ R and (b, c) $\in$ R, does it imply (a, c) $\in$ R?

Assume $(a, b) \in R$ and $(b, c) \in R$, where $a, b, c \in \mathbb{N}$.

$(a, b) \in R$ means $a = b^2$ ... (1)

$(b, c) \in R$ means $b = c^2$ ... (2)

For $(a, c)$ to be in R, we would need $a = c^2$.

Substitute equation (2) into equation (1):

$a = (c^2)^2$

$a = c^4$

We need to check if $a = c^4$ always implies $a = c^2$ for $a, b, c \in \mathbb{N}$ under the given conditions.

Let's construct an example: Let $c = 2 \in \mathbb{N}$.

From $b = c^2$, we get $b = 2^2 = 4$. Since $4 \in \mathbb{N}$, $(b, c) = (4, 2) \in R$.

From $a = b^2$, we get $a = 4^2 = 16$. Since $16 \in \mathbb{N}$, $(a, b) = (16, 4) \in R$.

So, we have $(16, 4) \in R$ and $(4, 2) \in R$.

Now we check if $(a, c) = (16, 2)$ is in R. This would require $16 = 2^2$, which is $16 = 4$. This is false.

So, $(a, b) \in R$ and $(b, c) \in R$ does not necessarily imply $(a, c) \in R$.

Justification: Take $a=16, b=4, c=2$. Since $16, 4, 2 \in \mathbb{N}$ and $16 = 4^2$, we have $(16, 4) \in R$. Since $4 = 2^2$, we have $(4, 2) \in R$. Now check if $(16, 2) \in R$. This would require $16 = 2^2$, which is $16 = 4$. This is false. Therefore, $(16, 2) \notin R$. Hence, $(a, b) \in R$ and $(b, c) \in R$ does not imply $(a, c) \in R$.

The statement (iii) is False.

Given:

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

Set B = {1, 5, 9, 11, 15, 16}

Relation candidate f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

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(i) Is f a relation from A to B?

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

Let's check each ordered pair in the given set f:

• For (1, 5): $1 \in A$ and $5 \in B$. This pair is in A $\times$ B.

• For (2, 9): $2 \in A$ and $9 \in B$. This pair is in A $\times$ B.

• For (3, 1): $3 \in A$ and $1 \in B$. This pair is in A $\times$ B.

• For (4, 5): $4 \in A$ and $5 \in B$. This pair is in A $\times$ B.

• For (2, 11): $2 \in A$ and $11 \in B$. This pair is in A $\times$ B.

Since every ordered pair in f has its first element from A and its second element from B, f is a subset of A $\times$ B.

Justification: All ordered pairs in the set f have their first component from set A and their second component from set B. Therefore, f is a subset of A $\times$ B by definition, which means f is a relation from A to B.

The statement (i) is True.

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(ii) Is f a function from A to B?

A relation f from set A to set B is a function if and only if every element in set A is related to exactly one element in set B.

This means two conditions must be met:

1. The domain of the relation f must be equal to set A.

2. No two distinct ordered pairs in f have the same first element.

Let's check these conditions for the given set f.

The domain of f is the set of all first elements of the ordered pairs: Domain(f) = {1, 2, 3, 4}. This is equal to set A = {1, 2, 3, 4}. So, condition 1 is met.

Now let's check condition 2 by looking at the first elements of the ordered pairs in f: (1, 5), (2, 9), (3, 1), (4, 5), (2, 11).

We observe that the element 2 from set A appears as the first element in two different ordered pairs: (2, 9) and (2, 11).

This means that the element 2 is related to two different elements in set B (9 and 11).

This violates the definition of a function, which requires each element in the domain to be mapped to a unique element in the codomain.

Justification: The element $2 \in A$ is associated with two different elements in B, namely 9 and 11, as shown by the ordered pairs (2, 9) $\in$ f and (2, 11) $\in$ f. A function must map each element of the domain to exactly one element of the codomain. Since 2 is mapped to both 9 and 11, f is not a function.

The statement (ii) is False.

The relation f is defined as $f = \{(ab, a + b) : a, b \in \mathbb{Z}\}$. This relation is a subset of $\mathbb{Z} \times \mathbb{Z}$.

For f to be a function from $\mathbb{Z}$ to $\mathbb{Z}$, every element in the domain of f must be mapped to exactly one element in the codomain $\mathbb{Z}$. The domain of f consists of all possible values of the product $ab$, where $a, b \in \mathbb{Z}$.

Let the input (the first component of the ordered pair) be $x = ab$, and the output (the second component) be $y = a + b$. For f to be a function, if we have two pairs of integers $(a_1, b_1)$ and $(a_2, b_2)$ such that their products are equal ($a_1 b_1 = a_2 b_2$), then their sums must also be equal ($a_1 + b_1 = a_2 + b_2$). If we can find a case where $a_1 b_1 = a_2 b_2$ but $a_1 + b_1 \neq a_2 + b_2$, then f is not a function.

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Let's try to find different pairs of integers $(a, b)$ that result in the same product $ab$. Consider the product $ab = 6$.

Case 1: Let $a = 1$ and $b = 6$.

The product is $ab = 1 \times 6 = 6$.

The sum is $a + b = 1 + 6 = 7$.

So, the ordered pair $(6, 7)$ is in f.

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Case 2: Let $a = 2$ and $b = 3.

The product is $ab = 2 \times 3 = 6$.

The sum is $a + b = 2 + 3 = 5$.

So, the ordered pair $(6, 5)$ is in f.

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We have found two ordered pairs in the relation f, which are $(6, 7)$ and $(6, 5)$. Both pairs have the same first component (6), but they have different second components (7 and 5). This means that the element 6 in the domain of f is mapped to two different elements (7 and 5) in the codomain $\mathbb{Z}$.

According to the definition of a function, each element in the domain must be mapped to exactly one element in the codomain. Since the element 6 is mapped to both 7 and 5, the relation f does not satisfy this condition.

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Justification: Consider the pairs of integers $(a_1, b_1) = (1, 6)$ and $(a_2, b_2) = (2, 3)$. Both $a_1, b_1, a_2, b_2$ are integers.

The product $a_1 b_1 = 1 \times 6 = 6$. The sum $a_1 + b_1 = 1 + 6 = 7$. Thus, $(6, 7) \in f$.

The product $a_2 b_2 = 2 \times 3 = 6$. The sum $a_2 + b_2 = 2 + 3 = 5$. Thus, $(6, 5) \in f$.

We have found that for the input value 6, the relation f gives two different output values, 7 and 5. This violates the definition of a function.

Therefore, f is not a function from $\mathbb{Z}$ to $\mathbb{Z}$.

The answer is No, f is not a function from $\mathbb{Z}$ to $\mathbb{Z}$.

Given:

Set A = {9, 10, 11, 12, 13}

Function f: A $\to$ $\mathbb{N}$ defined by f (n) = the highest prime factor of n.

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To Find:

The range of the function f.

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Solution:

The range of a function is the set of all possible output values for the elements in the domain. The domain of f is the set A = {9, 10, 11, 12, 13}. We need to find the highest prime factor for each element in A.

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For n = 9:

The prime factors of 9 are 3, 3.

The highest prime factor of 9 is 3.

So, $f(9) = 3$.

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For n = 10:

The prime factors of 10 are 2, 5.

The highest prime factor of 10 is 5.

So, $f(10) = 5$.

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For n = 11:

11 is a prime number.

The highest prime factor of 11 is 11.

So, $f(11) = 11$.

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For n = 12:

The prime factorization of 12 is $2 \times 2 \times 3 = 2^2 \times 3$.

The prime factors of 12 are 2, 3.

The highest prime factor of 12 is 3.

So, $f(12) = 3$.

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For n = 13:

13 is a prime number.

The highest prime factor of 13 is 13.

So, $f(13) = 13$.

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The set of output values obtained is $\{3, 5, 11, 3, 13\}$.

The range of f is the set of unique values in this set.

Range(f) = $\{3, 5, 11, 13\}$.

All elements in the range $\{3, 5, 11, 13\}$ are natural numbers, which is consistent with the codomain $\mathbb{N}$.

The range of f is $\{3, 5, 11, 13\}$.