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 $1 \in A$.

---

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

---

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:

The set A = {1, 2, 3, 4, 5, 6}.

A relation R is defined from set A to set A as R = {(x, y) : y = x + 1}.

First, let's find the ordered pairs that belong to the relation R by applying the rule $y = x + 1$ for each $x \in A$.

• If $x = 1$, then $y = 1 + 1 = 2$. Since $2 \in A$, the pair is (1, 2).
• If $x = 2$, then $y = 2 + 1 = 3$. Since $3 \in A$, the pair is (2, 3).
• If $x = 3$, then $y = 3 + 1 = 4$. Since $4 \in A$, the pair is (3, 4).
• If $x = 4$, then $y = 4 + 1 = 5$. Since $5 \in A$, the pair is (4, 5).
• If $x = 5$, then $y = 5 + 1 = 6$. Since $6 \in A$, the pair is (5, 6).
• If $x = 6$, then $y = 6 + 1 = 7$. Since $7 \notin A$, this pair is not in the relation R.

So, the relation R in roster form is R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.

---

(i) Depict this relation using an arrow diagram.

The arrow diagram consists of two ovals representing the set A (the domain and codomain). An arrow is drawn from each element $x$ in the first set to its corresponding element $y$ in the second set, according to the relation.

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

Domain: The domain of a relation is the set of all the first elements (or x-coordinates) of the ordered pairs in the relation.

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

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

Codomain: The codomain of a relation from set A to set B is the entire set B. Here, the relation is from A to A.

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

Range: The range of a relation is the set of all the second elements (or y-coordinates) of the ordered pairs in the relation.

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

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

By observing the given arrow diagram (Fig 2.6), we can identify the sets P and Q and the relationship between their elements.

Set P = {9, 4, 25}

Set Q = {5, 3, 2, 1, -2, -3, -5}

The arrows indicate the following pairings from P to Q:

• 9 is related to 3 and -3.
• 4 is related to 2 and -2.
• 25 is related to 5 and -5.

We can see a clear pattern: for each ordered pair $(x, y)$, where $x \in P$ and $y \in Q$, the element $x$ is the square of the element $y$. For example, $9 = 3^2$ and $9 = (-3)^2$.

---

(i) Set-builder form

The relation, let's call it R, can be described as the set of all ordered pairs $(x, y)$ such that $x$ is the square of $y$, where $x$ is an element of P and $y$ is an element of Q.

In set-builder notation, this is written as:

R = $\{(x, y) : x \text{ is the square of } y, x \in P, y \in Q\}$

Alternatively, using a mathematical equation:

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

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(ii) Roster form

The roster form is a list of all the ordered pairs $(x, y)$ that belong to the relation R. Based on the arrows in the diagram, we can write the relation as:

R = {(4, 2), (4, -2), (9, 3), (9, -3), (25, 5), (25, -5)}

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Domain and Range

Domain: The domain of the relation R is the set of all the first elements (or first components) of the ordered pairs in R.

Domain = {4, 9, 25}

Note that the domain is equal to the set P.

Range: The range of the relation R is the set of all the second elements (or second components) of the ordered pairs in R.

Range = {-5, -3, -2, 2, 3, 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 = $2^4$

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

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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.

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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) = $\{6, 7, 8\}$.

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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.

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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.

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(ii) Roster Form:

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

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

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(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.

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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)\}$

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(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 = $\{5, 6, 7, 8, 9, 10\}$.

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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.

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Roster Form:

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

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

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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

---

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

---

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 = $\{(x, y) : y = 2x, x, y \in N\}$.

---

To Find:

The domain of R.

The codomain of R.

The range of R.

Whether R is a function.

---

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:

A function $f: R \to R$ is defined by the rule $y = f(x) = x^2$, for all $x \in R$.

Completing the Table

We need to calculate the value of $y = f(x) = x^2$ for each given value of $x$.

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

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 and Range of the Function

Domain: The domain is the set of all possible input values (x-values) for which the function is defined. The function is given as $f: R \to R$, and the expression $x^2$ is defined for every real number $x$. Therefore, the domain of the function $f$ is the set of all real numbers.

Domain = R

Range: The range is the set of all possible output values (y-values) of the function. For any real number $x$, its square, $x^2$, is always non-negative (i.e., greater than or equal to zero). This is because the product of two negative numbers is positive, the product of two positive numbers is positive, and the square of zero is zero. Thus, the value of $y = f(x) = x^2$ will always be a non-negative real number.

Range = $[0, \infty)$ or $\{y \in R : y \ge 0\}$ (The set of all non-negative real numbers).

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Graph of the Function

The graph of the function $f(x) = x^2$ is a parabola that opens upwards. Its vertex is at the origin (0, 0). We can plot the points from the completed table to draw the curve: (-4, 16), (-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9), (4, 16).

Given:

The function is $f: R \to R$ defined by the rule $f(x) = x^3$, for all $x \in R$.

To draw the graph of this function, it is helpful to first create a table of values by selecting some values for $x$ and calculating the corresponding values for $y = f(x)$.

Table of Values

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

• If $x = -2$, $y = (-2)^3 = -8$
• If $x = -1$, $y = (-1)^3 = -1$
• If $x = 0$, $y = (0)^3 = 0$
• If $x = 1$, $y = (1)^3 = 1$
• If $x = 2$, $y = (2)^3 = 8$

These points can be summarized in a table:

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

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Domain and Range

Domain: The function $f(x) = x^3$ is defined for all real numbers. Therefore, the Domain of f is R.

Range: For any real number $y$, there exists a real number $x$ such that $x = \sqrt[3]{y}$. This means that the output of the function can be any real number (positive, negative, or zero). Therefore, the Range of f is R.

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Graph of the Function

By plotting the points from the table, such as (–2, –8), (–1, –1), (0, 0), (1, 1), and (2, 8), and connecting them with a smooth curve, we obtain the graph of the function $f(x) = x^3$. The graph passes through the origin. For positive values of $x$, the graph is in the first quadrant, and for negative values of $x$, the graph is in the third quadrant.

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 are defined as:

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

$g(x) = x$

Both functions are defined for the set of non-negative real numbers. So, the domain of both $f$ and $g$ is $[0, \infty)$.

The domain of the sum, difference, and product of these functions will be the intersection of their individual domains, which is $[0, \infty) \cap [0, \infty) = [0, \infty)$.

The domain of the quotient will be the intersection of their domains, excluding points where the denominator is zero.

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(i) Sum of the functions: $(f + g)(x)$

By definition, $(f + g)(x) = f(x) + g(x)$.

Substituting the given functions:

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

The domain of this function is $[0, \infty)$.

Therefore, $(f + g)(x) = \sqrt{x} + x$, for $x \in [0, \infty)$.

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(ii) Difference of the functions: $(f – g)(x)$

By definition, $(f – g)(x) = f(x) – g(x)$.

Substituting the given functions:

$(f – g)(x) = \sqrt{x} - x$

The domain of this function is $[0, \infty)$.

Therefore, $(f – g)(x) = \sqrt{x} - x$, for $x \in [0, \infty)$.

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(iii) Product of the functions: $(fg)(x)$

By definition, $(fg)(x) = f(x) \cdot g(x)$.

Substituting the given functions:

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

Using the laws of exponents, where $\sqrt{x} = x^{1/2}$ and $x = x^1$:

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

The domain of this function is $[0, \infty)$.

Therefore, $(fg)(x) = x^{3/2}$, for $x \in [0, \infty)$.

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(iv) Quotient of the functions: $\left(\frac{f}{g}\right)(x)$

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

Substituting the given functions:

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

The domain is the intersection of the domains of $f$ and $g$, which is $[0, \infty)$, but we must exclude the values of $x$ for which $g(x) = 0$. Here, $g(x) = x = 0$ when $x=0$. So, we must exclude $x=0$ from the domain.

The domain of this function is $(0, \infty)$.

Simplifying the expression for $x > 0$:

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

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

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.

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

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

$t(0) = 0 + 32$

$t(0) = 32$

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

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

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

$t(28) = 50.4 + 32$

$t(28) = 82.4$

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(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 $\{y \in \mathbb{R} \mid y < 2\}$.

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(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 $\{y \in \mathbb{R} \mid y \ge 2\}$.

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(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 $(-\infty, \infty)$.

Given:

The real function is defined as $f: R \to R$ by the rule $f(x) = x + 10$.

This is a linear function of the form $y = mx + c$, where the slope $m=1$ and the y-intercept $c=10$. The graph of this function will be a straight line.

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Finding Points for the Graph

To sketch the graph of a straight line, we need to find at least two points that lie on the line. We can create a table of values for $x$ and the corresponding $y = f(x)$.

• When $x = 0$, $y = 0 + 10 = 10$. The point is (0, 10). This is the y-intercept.
• When $x = -10$, $y = -10 + 10 = 0$. The point is (-10, 0). This is the x-intercept.
• When $x = -5$, $y = -5 + 10 = 5$. The point is (-5, 5).

The table of values is as follows:

$x$	$y = f(x) = x + 10$	Point (x, y)
-10	0	(-10, 0)
-5	5	(-5, 5)
0	10	(0, 10)

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Domain and Range

Domain: The function $f(x) = x + 10$ is defined for all real numbers $x$. Therefore, the domain of the function is the set of all real numbers, R.

Range: For any real number $y$, we can find a real number $x$ such that $y = x + 10$ (specifically, $x = y - 10$). This means the output can be any real number. Therefore, the range of the function is the set of all real numbers, R.

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Sketching the Graph

The graph is a straight line that passes through the points we calculated, such as (-10, 0) and (0, 10). We plot these points on a Cartesian plane and draw a straight line through them.

Given:

A relation R is defined on the set of rational numbers, Q, as:

R = $\{(a, b) : a, b \in Q \text{ and } a – b \in Z\}$, where Q is the set of rational numbers and Z is the set of integers.

This means that an ordered pair $(a, b)$ belongs to the relation R if both $a$ and $b$ are rational numbers and their difference is an integer.

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(i) To show that $(a, a) \in R$ for all $a \in Q$ (Reflexive Property)

Proof:

Let $a$ be any rational number, so $a \in Q$.

To check if $(a, a) \in R$, we need to verify if the condition $a - a \in Z$ is true.

Consider the difference $a - a$.

$a - a = 0$

Since 0 is an integer (i.e., $0 \in Z$), the condition for the relation is satisfied.

Therefore, $(a, a) \in R$ for all $a \in Q$.

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

Proof:

Let's assume that $(a, b) \in R$.

By the definition of the relation R, this means that $a, b \in Q$ and $a - b \in Z$.

Let $a - b = k$, where $k$ is some integer ($k \in Z$).

Now, we need to check if $(b, a) \in R$. For this to be true, the difference $b - a$ must be an integer.

Consider the difference $b - a$.

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

Since we assumed $a - b = k$, we can substitute this into the equation:

$b - a = -k$

Because $k$ is an integer, its negative, $-k$, is also an integer.

So, $b - a \in Z$.

This satisfies the condition for $(b, a)$ to be in R.

Therefore, if $(a, b) \in R$, then $(b, a) \in R$.

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

Proof:

Let's assume that $(a, b) \in R$ and $(b, c) \in R$.

From $(a, b) \in R$, we know that $a, b \in Q$ and $a - b \in Z$.

Let $a - b = k_1$, where $k_1$ is an integer.

From $(b, c) \in R$, we know that $b, c \in Q$ and $b - c \in Z$.

Let $b - c = k_2$, where $k_2$ is an integer.

Now, we need to check if $(a, c) \in R$. For this to be true, the difference $a - c$ must be an integer.

We can write the expression $a - c$ as:

$a - c = (a - b) + (b - c)$

Substituting the integer values we defined:

$a - c = k_1 + k_2$

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

So, $a - c \in Z$.

This satisfies the condition for $(a, c)$ to be in R.

Therefore, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Given:

The function $f$ is a linear function from the set of integers (Z) to the set of integers (Z).

The function is given by the set of ordered pairs: $f = \{(1, 1), (2, 3), (0, –1), (–1, –3)\}$.

This means that:

• $f(1) = 1$
• $f(2) = 3$
• $f(0) = -1$
• $f(-1) = -3$

To Find:

The expression for the function $f(x)$.

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Solution:

Since $f$ is a linear function, its general form can be written as:

$f(x) = mx + c$

where $m$ is the slope and $c$ is the y-intercept. Our goal is to find the values of $m$ and $c$.

We can use any two of the given ordered pairs to form a system of linear equations and solve for $m$ and $c$.

Using the point $(0, -1)$ is the most straightforward way to find $c$, as it represents the y-intercept.

Substitute $x=0$ and $f(x)=-1$ into the equation:

$f(0) = m(0) + c$

$-1 = 0 + c$

$c = -1$

Now we know the function is of the form $f(x) = mx - 1$. We can use another point, for example $(1, 1)$, to find the value of $m$.

Substitute $x=1$ and $f(x)=1$ into the equation:

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

$1 = m - 1$

$m = 1 + 1$

$m = 2$

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

Verification:

Let's check if this function holds true for the other two points:

For the point $(2, 3)$:

$f(2) = 2(2) - 1 = 4 - 1 = 3$. This is correct.

For the point $(-1, -3)$:

$f(-1) = 2(-1) - 1 = -2 - 1 = -3$. This is also correct.

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

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Alternate Solution:

We can pick any two points, say $(1, 1)$ and $(2, 3)$, and form a system of two equations.

For $(1, 1)$: $f(1) = m(1) + c \implies 1 = m + c$

For $(2, 3)$: $f(2) = m(2) + c \implies 3 = 2m + c$

Subtracting equation (i) from equation (ii):

$(2m + c) - (m + c) = 3 - 1$

$m = 2$

Substitute the value of $m=2$ back into equation (i):

$2 + c = 1$

$c = 1 - 2$

$c = -1$

This gives the same result. The linear function is $f(x) = 2x - 1$.

Given:

The function is defined as:

$f(x) = \frac{x^{2} + 3x + 5}{x^{2} - 5x + 4}$

To Find:

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

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Solution:

The given function is a rational function, which is a fraction of two polynomials. A rational function is defined for all real numbers except for those values of $x$ that make the denominator equal to zero, as division by zero is undefined.

To find the domain, we must first find the values of $x$ for which the denominator is zero. We set the denominator equal to zero:

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

This is a quadratic equation which can be solved by factorization (splitting the middle term).

$x^2 - 4x - 1x + 4 = 0$

Take $x$ common from the first two terms and $-1$ common from the last two terms:

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

Now, take $(x - 4)$ common:

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

This equation is true if either $(x - 4) = 0$ or $(x - 1) = 0$.

If $x - 4 = 0$, then $x = 4$.

If $x - 1 = 0$, then $x = 1$.

Thus, the denominator of the function $f(x)$ becomes zero when $x=1$ or $x=4$. Therefore, the function is not defined at these two points.

The domain of the function is the set of all real numbers except 1 and 4.

This can be expressed in set notation as $R - \{1, 4\}$.

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

Domain = $R - \{1, 4\}$

Given:

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

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

To draw the graph, we will analyze and plot each piece of the function separately.

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Analysis for Graphing

The graph will be composed of three parts corresponding to the three conditions on $x$.

Part 1: For $x < 0$, the function is $f(x) = 1 - x$.

This is a linear function, so its graph is a straight line. Let's find some points on this line for negative values of $x$.

• If $x = -1$, $y = 1 - (-1) = 2$. Point: (-1, 2).
• If $x = -2$, $y = 1 - (-2) = 3$. Point: (-2, 3).
• If $x = -3$, $y = 1 - (-3) = 4$. Point: (-3, 4).

As $x$ approaches 0 from the left, $f(x)$ approaches $1 - 0 = 1$. So, this part of the graph is a ray starting from just to the left of the point (0, 1) and extending upwards to the left.

Part 2: For $x = 0$, the function is $f(x) = 1$.

This defines a single point on the graph, which is (0, 1). This point will be plotted as a solid dot at (0, 1).

Part 3: For $x > 0$, the function is $f(x) = x + 1$.

This is also a linear function, so its graph is a straight line. Let's find some points on this line for positive values of $x$.

• If $x = 1$, $y = 1 + 1 = 2$. Point: (1, 2).
• If $x = 2$, $y = 2 + 1 = 3$. Point: (2, 3).
• If $x = 3$, $y = 3 + 1 = 4$. Point: (3, 4).

As $x$ approaches 0 from the right, $f(x)$ approaches $0 + 1 = 1$. So, this part of the graph is a ray starting from just to the right of the point (0, 1) and extending upwards to the right.

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Drawing the Graph

We combine the three parts to create the final graph:

1. Plot the line for $y = 1 - x$ in the second quadrant (where $x < 0$).

2. Plot the line for $y = x + 1$ in the first quadrant (where $x > 0$).

3. Place a solid dot at the point (0, 1), which serves as the vertex where the two lines meet.

The resulting graph is a 'V' shape with its vertex at (0, 1). This function is actually equivalent to the modulus function shifted up by one unit, i.e., $f(x) = |x| + 1$.

A relation is considered a function if every element in its domain has exactly one and only one image (output value).

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Analysis of the relation f

The relation f is defined as:

$f(x) = \begin{cases} x^2 &,& 0 \leq x \leq 3 \\ 3x &,& 3 \leq x \leq 10 \end{cases}$

The domain of this relation is the union of the two intervals, which is $[0, 3] \cup [3, 10] = [0, 10]$.

For any value of $x$ such that $0 \leq x < 3$, the value of $f(x)$ is uniquely defined as $x^2$.

For any value of $x$ such that $3 < x \leq 10$, the value of $f(x)$ is uniquely defined as $3x$.

The only point where the definition could be ambiguous is at the boundary point, $x = 3$. We need to check if the value of $f(3)$ is unique.

According to the first piece ($f(x) = x^2$):

$f(3) = (3)^2 = 9$

According to the second piece ($f(x) = 3x$):

$f(3) = 3(3) = 9$

Since both definitions give the same value for $f(3)$, the element $x=3$ has a unique image, which is 9.

Because every element $x$ in the domain $[0, 10]$ has exactly one image, the relation f is a function.

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Analysis of the relation g

The relation g is defined as:

$g(x) = \begin{cases} x^2 &,& 0 \leq x \leq 2 \\ 3x &,& 2 \leq x \leq 10 \end{cases}$

The domain of this relation is the union of the two intervals, which is $[0, 2] \cup [2, 10] = [0, 10]$.

Let's check the boundary point, $x = 2$, to see if it has a unique image.

According to the first piece ($g(x) = x^2$):

$g(2) = (2)^2 = 4$

According to the second piece ($g(x) = 3x$):

$g(2) = 3(2) = 6$

Here, for the single input value $x = 2$, the relation g gives two different output values, 4 and 6. This means the element 2 has two different images.

Since one element in the domain has more than one image, the relation violates the fundamental definition of a function.

Therefore, the relation g is not a function.

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

We need to find the value of $\frac{f(1.1) - f(1)}{(1.1 - 1)}$.

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First, let's find the value of $f(1.1)$ by substituting $x = 1.1$ into the function:

$f(1.1) = (1.1)^2$

$f(1.1) = 1.21$

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Next, let's find the value of $f(1)$ by substituting $x = 1$ into the function:

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

$f(1) = 1$

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Now, let's calculate the numerator $f(1.1) - f(1)$:

$f(1.1) - f(1) = 1.21 - 1$

$f(1.1) - f(1) = 0.21$

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Next, let's calculate the denominator $1.1 - 1$:

$1.1 - 1 = 0.1$

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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:

$\frac{0.21 \times 10}{0.1 \times 10} = \frac{2.1}{1}$

$\frac{0.21}{0.1} = 2.1$

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Thus, the value of $\frac{f(1.1) - f(1)}{(1.1 - 1)}$ is 2.1.

The given function is $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

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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.

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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:

$x = 2 \quad \text{or} \quad x = 6$

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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 $f(x) = \sqrt{x - 1}$.

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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 $[1, \infty)$.

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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 $[0, \infty)$.

Given:

The real function is defined as $f(x) = |x - 1|$.

This is the absolute value function, which gives the non-negative value of the expression inside it.

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Domain of the function

The domain is the set of all possible input values of $x$ for which the function is defined.

The expression inside the absolute value, $x - 1$, is a simple polynomial, which is defined for all real numbers $x$.

The absolute value operation is also defined for any real number.

Therefore, there are no restrictions on the value of $x$. The function $f(x) = |x - 1|$ is defined for every real number.

The domain of the function is the set of all real numbers, denoted by R.

Domain = R

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Range of the function

The range is the set of all possible output values of $f(x)$.

By the definition of the absolute value (or modulus), the output is always non-negative. That is, for any real number $a$, $|a| \ge 0$.

In our case, $f(x) = |x - 1|$. Therefore, the value of $f(x)$ must be greater than or equal to zero.

$f(x) \ge 0$ for all $x \in R$.

This means the range is the set of all non-negative real numbers.

This can be written as the interval $[0, \infty)$ or in set-builder notation as $\{y \in R : y \ge 0\}$.

Range = $[0, \infty)$

Given:

The function is defined as $f: R \to R$ with the rule:

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

The domain of the function is the set of all real numbers, R.

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. Let $y$ be an element in the range of $f$.

Then, $y = f(x) = \frac{x^2}{1 + x^2}$ for some $x \in R$.

We can find the range by expressing $x$ in terms of $y$.

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

$y + yx^2 = x^2$

$y = x^2 - yx^2$

$y = x^2(1 - y)$

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

Since $x$ is a real number, its square, $x^2$, must be non-negative. Therefore, $x^2 \ge 0$.

This implies that:

$\frac{y}{1 - y} \ge 0$

This inequality holds true under two conditions:

Case 1: $y \ge 0$ and $1 - y > 0$.

This means $y \ge 0$ and $y < 1$. Combining these gives $0 \le y < 1$.

Case 2: $y \le 0$ and $1 - y < 0$.

This means $y \le 0$ and $y > 1$. There is no value of $y$ that can satisfy both conditions simultaneously.

Therefore, the only possible values for $y$ are in the interval $[0, 1)$.

The range of the function $f$ is the set of all real numbers from 0 (inclusive) to 1 (exclusive).

Range of f = $[0, 1)$

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Alternate Solution (By Analysis)

Let's analyze the function $f(x) = \frac{x^2}{1 + x^2}$.

1. Finding the lower bound of the range:

For any real number $x$, the numerator $x^2$ is always non-negative, i.e., $x^2 \ge 0$.

The denominator $1 + x^2$ is always positive, i.e., $1 + x^2 > 0$.

Since the numerator is non-negative and the denominator is positive, the value of the function $f(x)$ must be non-negative.

$f(x) \ge 0$.

The value $f(x) = 0$ is achieved when $x = 0$, so 0 is included in the range.

2. Finding the upper bound of the range:

We can rewrite the function as:

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

We know that $x^2 \ge 0$, which implies $1 + x^2 \ge 1$.

Taking the reciprocal reverses the inequality: $0 < \frac{1}{1 + x^2} \le 1$.

Multiplying by -1 reverses the inequality again: $-1 \le -\frac{1}{1 + x^2} < 0$.

Adding 1 to all parts of the inequality:

$1 - 1 \le 1 - \frac{1}{1 + x^2} < 1 - 0$

$0 \le f(x) < 1$

From this analysis, we can see that the value of $f(x)$ is always greater than or equal to 0 and strictly less than 1.

Therefore, the range of the function is the interval from 0 to 1, including 0 but not 1.

Range of f = $[0, 1)$

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 $\{3, 5, 11, 13\}$.