Relations and Functions

Given: Two ordered pairs are equal: $\left(\dfrac{x}{3} + 1,\ y - \dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right)$.

Concept: Two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if $a = c$ and $b = d$.

Equating first elements:
$$\frac{x}{3} + 1 = \frac{5}{3}$$
$$\frac{x}{3} = \frac{5}{3} - 1 = \frac{5-3}{3} = \frac{2}{3}$$
$$x = 2$$

Equating second elements:
$$y - \frac{2}{3} = \frac{1}{3}$$
$$y = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1$$

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

Given: $n(\mathrm{A}) = 3$ and $\mathrm{B} = \{3, 4, 5\}$, so $n(\mathrm{B}) = 3$.

Concept: If $n(\mathrm{A}) = p$ and $n(\mathrm{B}) = q$, then $n(\mathrm{A} \times \mathrm{B}) = p \times q$.

Calculation:
$$n(\mathrm{A} \times \mathrm{B}) = n(\mathrm{A}) \times n(\mathrm{B}) = 3 \times 3 = 9$$

Answer: The number of elements in $\mathrm{A} \times \mathrm{B}$ is $\mathbf{9}$.

Given: $\mathrm{G} = \{7, 8\}$ and $\mathrm{H} = \{5, 4, 2\}$.

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

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

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

(i) False.

$\mathrm{P} = \{m, n\}$ and $\mathrm{Q} = \{n, m\} = \{m, n\}$.

The correct Cartesian product is:
$$\mathrm{P} \times \mathrm{Q} = \{(m,m),\ (m,n),\ (n,m),\ (n,n)\}$$

The given statement lists only 2 pairs instead of 4, so it is False.

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

(ii) True.

By definition, if A and B are non-empty sets, $\mathrm{A} \times \mathrm{B}$ is indeed a non-empty set of ordered pairs $(x, y)$ with $x \in \mathrm{A}$ and $y \in \mathrm{B}$.

(iii) True.

$\mathrm{B} \cap \phi = \phi$ (intersection of any set with the empty set is empty).

Therefore, $\mathrm{A} \times (\mathrm{B} \cap \phi) = \mathrm{A} \times \phi = \phi$.

Given: $\mathrm{A} = \{-1, 1\}$.

Concept: $\mathrm{A} \times \mathrm{A} \times \mathrm{A} = \{(a, b, c) : a, b, c \in \mathrm{A}\}$.

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

There are $2^3 = 8$ elements in total.

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

Concept: In a Cartesian product, A is the set of all first elements and B is the set of all second elements of the ordered pairs.

Finding A: Set of all first elements $= \{a, b\}$
$$\therefore \mathrm{A} = \{a, b\}$$

Finding B: Set of all second elements $= \{x, y\}$
$$\therefore \mathrm{B} = \{x, y\}$$

Given: $\mathrm{A} = \{1,2\}$, $\mathrm{B} = \{1,2,3,4\}$, $\mathrm{C} = \{5,6\}$, $\mathrm{D} = \{5,6,7,8\}$.

(i) Verify $\mathrm{A} \times (\mathrm{B} \cap \mathrm{C}) = (\mathrm{A} \times \mathrm{B}) \cap (\mathrm{A} \times \mathrm{C})$:

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

RHS:
$$\mathrm{A} \times \mathrm{B} = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$$
$$\mathrm{A} \times \mathrm{C} = \{(1,5),(1,6),(2,5),(2,6)\}$$
$$(\mathrm{A} \times \mathrm{B}) \cap (\mathrm{A} \times \mathrm{C}) = \phi \quad (\text{no common elements})$$

Since LHS $= \phi =$ RHS, the statement is verified.

(ii) Verify $\mathrm{A} \times \mathrm{C} \subseteq \mathrm{B} \times \mathrm{D}$:

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

Checking each element of $\mathrm{A} \times \mathrm{C}$:
- $(1,5) \in \mathrm{B} \times \mathrm{D}$ ✓ (since $1 \in \mathrm{B}$, $5 \in \mathrm{D}$)
- $(1,6) \in \mathrm{B} \times \mathrm{D}$ ✓
- $(2,5) \in \mathrm{B} \times \mathrm{D}$ ✓
- $(2,6) \in \mathrm{B} \times \mathrm{D}$ ✓

Since every element of $\mathrm{A} \times \mathrm{C}$ belongs to $\mathrm{B} \times \mathrm{D}$, we have $\mathrm{A} \times \mathrm{C} \subseteq \mathrm{B} \times \mathrm{D}$. Verified.

Given: $\mathrm{A} = \{1,2\}$, $\mathrm{B} = \{3,4\}$.

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

Number of subsets: $n(\mathrm{A} \times \mathrm{B}) = 4$, so the number of subsets $= 2^4 = \mathbf{16}$.

List of all subsets:
1. $\phi$
2. $\{(1,3)\}$
3. $\{(1,4)\}$
4. $\{(2,3)\}$
5. $\{(2,4)\}$
6. $\{(1,3),(1,4)\}$
7. $\{(1,3),(2,3)\}$
8. $\{(1,3),(2,4)\}$
9. $\{(1,4),(2,3)\}$
10. $\{(1,4),(2,4)\}$
11. $\{(2,3),(2,4)\}$
12. $\{(1,3),(1,4),(2,3)\}$
13. $\{(1,3),(1,4),(2,4)\}$
14. $\{(1,3),(2,3),(2,4)\}$
15. $\{(1,4),(2,3),(2,4)\}$
16. $\{(1,3),(1,4),(2,3),(2,4)\}$

Given: $n(\mathrm{A}) = 3$, $n(\mathrm{B}) = 2$, and $(x,1),\ (y,2),\ (z,1) \in \mathrm{A} \times \mathrm{B}$, where $x, y, z$ are distinct.

Finding A: The first elements of the ordered pairs in $\mathrm{A} \times \mathrm{B}$ belong to A.
First elements are $x, y, z$ (all distinct), and $n(\mathrm{A}) = 3$.
$$\therefore \mathrm{A} = \{x, y, z\}$$

Finding B: The second elements of the ordered pairs in $\mathrm{A} \times \mathrm{B}$ belong to B.
Second elements are $1, 2, 1$, i.e., the distinct values are $1$ and $2$, and $n(\mathrm{B}) = 2$.
$$\therefore \mathrm{B} = \{1, 2\}$$

Given: $n(\mathrm{A} \times \mathrm{A}) = 9$ and $(-1, 0),\ (0, 1) \in \mathrm{A} \times \mathrm{A}$.

Finding $n(\mathrm{A})$:
$$n(\mathrm{A} \times \mathrm{A}) = [n(\mathrm{A})]^2 = 9 \implies n(\mathrm{A}) = 3$$

Finding A: Since $(-1, 0) \in \mathrm{A} \times \mathrm{A}$, both $-1$ and $0$ must be in A. Since $(0, 1) \in \mathrm{A} \times \mathrm{A}$, $1$ must also be in A. We already have 3 distinct elements $\{-1, 0, 1\}$ and $n(\mathrm{A}) = 3$.
$$\therefore \mathrm{A} = \{-1, 0, 1\}$$

Finding all elements of $\mathrm{A} \times \mathrm{A}$:
$$\mathrm{A} \times \mathrm{A} = \{(-1,-1),\ (-1,0),\ (-1,1),\ (0,-1),\ (0,0),\ (0,1),\ (1,-1),\ (1,0),\ (1,1)\}$$

Remaining elements (other than $(-1,0)$ and $(0,1)$) are:
$$\{(-1,-1),\ (-1,1),\ (0,-1),\ (0,0),\ (1,-1),\ (1,0),\ (1,1)\}$$

Given: $\mathrm{A} = \{1, 2, 3, \ldots, 14\}$ and $\mathrm{R} = \{(x, y) : 3x - y = 0,\ x, y \in \mathrm{A}\}$.

Condition: $3x - y = 0 \Rightarrow y = 3x$.

We need $x, y \in \mathrm{A}$, i.e., both $x$ and $3x$ must lie in $\{1, 2, \ldots, 14\}$.

| $x$ | $y = 3x$ | In A? |
|-----|----------|-------|
| 1 | 3 | Yes |
| 2 | 6 | Yes |
| 3 | 9 | Yes |
| 4 | 12 | Yes |
| 5 | 15 | No (15 > 14) |

So the relation in roster form is:
$$\mathrm{R} = \{(1,3),\ (2,6),\ (3,9),\ (4,12)\}$$

Domain = set of first elements $= \{1, 2, 3, 4\}$

Codomain = A $= \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$

Range = set of second elements $= \{3, 6, 9, 12\}$

Given: \mathrm{R} = \{(x, y) : y = x + 5,\ x \in \mathbf{N},\ x < 4\}.

Natural numbers less than 4: $x = 1, 2, 3$.

| $x$ | $y = x + 5$ |
|-----|-------------|
| 1 | 6 |
| 2 | 7 |
| 3 | 8 |

Roster form:
$$\mathrm{R} = \{(1, 6),\ (2, 7),\ (3, 8)\}$$

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

Range $= \{6, 7, 8\}$

Given: $\mathrm{A} = \{1, 2, 3, 5\}$, $\mathrm{B} = \{4, 6, 9\}$.

Condition: $|x - y|$ is odd (i.e., $x - y$ is odd), which happens when one of $x, y$ is even and the other is odd.

Checking all pairs $(x, y)$ with $x \in \mathrm{A}$, $y \in \mathrm{B}$:

- $x=1$ (odd): $y=4$ (even) → $1-4=-3$ odd ✓; $y=6$ (even) → odd ✓; $y=9$ (odd) → even ✗
- $x=2$ (even): $y=4$ (even) → even ✗; $y=6$ (even) → even ✗; $y=9$ (odd) → odd ✓
- $x=3$ (odd): $y=4$ → odd ✓; $y=6$ → odd ✓; $y=9$ → even ✗
- $x=5$ (odd): $y=4$ → odd ✓; $y=6$ → odd ✓; $y=9$ → even ✗

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

Note: The figure (Fig 2.7) is not visible in the OCR text. Based on the standard NCERT textbook, Fig 2.7 shows the relation where elements of P = {5, 6, 7} are related to elements of Q = {3, 4, 5} by the rule $x - y = 2$ (i.e., $y = x - 2$), giving pairs (5,3), (6,4), (7,5).

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

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

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

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

Given: $\mathrm{A} = \{1, 2, 3, 4, 6\}$, $\mathrm{R} = \{(a,b) : a,b \in \mathrm{A},\ a \mid b\}$.

(i) Roster form: We list all pairs $(a, b)$ where $b$ is exactly divisible by $a$:

- $a=1$: $b$ can be $1, 2, 3, 4, 6$ → $(1,1),(1,2),(1,3),(1,4),(1,6)$
- $a=2$: $b$ can be $2, 4, 6$ → $(2,2),(2,4),(2,6)$
- $a=3$: $b$ can be $3, 6$ → $(3,3),(3,6)$
- $a=4$: $b$ can be $4$ → $(4,4)$
- $a=6$: $b$ can be $6$ → $(6,6)$

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

(ii) Domain = set of all first elements $= \{1, 2, 3, 4, 6\}$

(iii) Range = set of all second elements $= \{1, 2, 3, 4, 6\}$

Given: $\mathrm{R} = \{(x, x+5) : x \in \{0,1,2,3,4,5\}\}$.

Listing all ordered pairs:

| $x$ | $x+5$ |
|-----|-------|
| 0 | 5 |
| 1 | 6 |
| 2 | 7 |
| 3 | 8 |
| 4 | 9 |
| 5 | 10 |

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

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

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

Given: $\mathrm{R} = \{(x, x^3) : x \text{ is a prime number less than } 10\}$.

Prime numbers less than 10: $2, 3, 5, 7$.

| $x$ | $x^3$ |
|-----|-------|
| 2 | 8 |
| 3 | 27 |
| 5 | 125 |
| 7 | 343 |

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

Given: $\mathrm{A} = \{x, y, z\}$, $\mathrm{B} = \{1, 2\}$.

$n(\mathrm{A}) = 3$, $n(\mathrm{B}) = 2$.

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

Concept: The number of relations from A to B = number of subsets of $\mathrm{A} \times \mathrm{B} = 2^{n(\mathrm{A} \times \mathrm{B})}$.

$$\text{Number of relations} = 2^6 = \mathbf{64}$$

Given: $\mathrm{R} = \{(a, b) : a, b \in \mathbf{Z},\ a - b \in \mathbf{Z}\}$.

Observation: For any two integers $a$ and $b$, their difference $a - b$ is always an integer. Therefore, every pair $(a, b)$ with $a, b \in \mathbf{Z}$ satisfies the condition.

This means $\mathrm{R} = \mathbf{Z} \times \mathbf{Z}$.

Domain = set of all first elements $= \mathbf{Z}$

Range = set of all second elements $= \mathbf{Z}$

Concept: A relation is a function if every element in the domain has exactly one image (no first element is repeated with different second elements).

(i) $\{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\}$

Each first element $2, 5, 8, 11, 14, 17$ appears exactly once. So this is a function.

- Domain $= \{2, 5, 8, 11, 14, 17\}$
- Range $= \{1\}$

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

Each first element $2, 4, 6, 8, 10, 12, 14$ appears exactly once. So this is a function.

- Domain $= \{2, 4, 6, 8, 10, 12, 14\}$
- Range $= \{1, 2, 3, 4, 5, 6, 7\}$

(iii) $\{(1,3),(1,5),(2,5)\}$

The element $1$ appears as the first element in two pairs: $(1,3)$ and $(1,5)$, with two different images $3$ and $5$. So this is NOT a function.

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

Domain: $|x|$ is defined for all real numbers.
$$\text{Domain} = \mathbf{R}$$

Range: For any $x \in \mathbf{R}$, $|x| \geq 0$, so $-|x| \leq 0$.
The function takes all values $\leq 0$.
$$\text{Range} = (-\infty, 0] = \{y \in \mathbf{R} : y \leq 0\}$$

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

Domain: The expression under the square root must be non-negative:
$$9 - x^2 \geq 0 \implies x^2 \leq 9 \implies -3 \leq x \leq 3$$
$$\text{Domain} = [-3, 3]$$

Range: When $x \in [-3, 3]$, $x^2 \in [0, 9]$, so $9 - x^2 \in [0, 9]$, and $\sqrt{9-x^2} \in [0, 3]$.
$$\text{Range} = [0, 3]$$

Given: $f(x) = 2x - 5$.

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

(ii) $f(7)$:
$$f(7) = 2(7) - 5 = 14 - 5 = 9$$

(iii) $f(-3)$:
$$f(-3) = 2(-3) - 5 = -6 - 5 = -11$$

Given: $t(\mathrm{C}) = \dfrac{9\mathrm{C}}{5} + 32$.

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

(ii) $t(28)$:
$$t(28) = \frac{9 \times 28}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4$$

(iii) $t(-10)$:
$$t(-10) = \frac{9 \times (-10)}{5} + 32 = \frac{-90}{5} + 32 = -18 + 32 = 14$$

(iv) Value of $C$ when $t(C) = 212$:
$$\frac{9C}{5} + 32 = 212$$
$$\frac{9C}{5} = 212 - 32 = 180$$
$$9C = 900$$
$$C = 100$$

Answer: $t(0) = 32$, $t(28) = 82.4$, $t(-10) = 14$, and $C = 100$ when $t(C) = 212$.

(i) f(x) = 2 - 3x,\ x \in \mathbf{R},\ x > 0:

Since x > 0, we have 3x > 0, so -3x < 0, thus 2 - 3x < 2.
As $x \to 0^+$, $f(x) \to 2$ (but $x = 0$ is excluded, so $f(x)$ approaches but never equals 2).
As $x \to +\infty$, $f(x) \to -\infty$.

\text{Range} = (-\infty, 2) = \{y \in \mathbf{R} : y < 2\}

(ii) $f(x) = x^2 + 2,\ x \in \mathbf{R}$:

For all real $x$, $x^2 \geq 0$, so $x^2 + 2 \geq 2$.
The minimum value is $2$ (attained at $x = 0$), and $f(x)$ can be arbitrarily large.

$$\text{Range} = [2, +\infty) = \{y \in \mathbf{R} : y \geq 2\}$$

(iii) $f(x) = x,\ x \in \mathbf{R}$:

This is the identity function. For every real number $y$, there exists $x = y$ such that $f(x) = y$.

$$\text{Range} = \mathbf{R}$$

For $f$ to be a function: Every element in the domain must have a unique image.

The domain of $f$ is $[0, 10]$. The two pieces overlap only at $x = 3$.

At $x = 3$: From the first piece, $f(3) = 3^2 = 9$. From the second piece, $f(3) = 3 \times 3 = 9$.

Both pieces give the same value at $x = 3$. For all other $x$, only one piece applies. Therefore, every element in $[0, 10]$ has a unique image.

$$\therefore f \text{ is a function.}$$

For $g$: The two pieces overlap at $x = 2$.

At $x = 2$: From the first piece, $g(2) = 2^2 = 4$. From the second piece, $g(2) = 3 \times 2 = 6$.

The two pieces give different values ($4 \neq 6$) at $x = 2$, so the element $2$ in the domain has two different images.

$$\therefore g \text{ is not a function.}$$

Given: $f(x) = x^2$.

$$f(1.1) = (1.1)^2 = 1.21$$
$$f(1) = (1)^2 = 1$$

$$\frac{f(1.1) - f(1)}{1.1 - 1} = \frac{1.21 - 1}{0.1} = \frac{0.21}{0.1} = \mathbf{2.1}$$

Given: $f(x) = \dfrac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

Concept: The function is defined for all real $x$ except where the denominator is zero.

Setting denominator $= 0$:
$$x^2 - 8x + 12 = 0$$
$$(x - 6)(x - 2) = 0$$
$$x = 6 \quad \text{or} \quad x = 2$$

The function is undefined at $x = 2$ and $x = 6$.

$$\text{Domain} = \mathbf{R} \setminus \{2, 6\} = \{x \in \mathbf{R} : x \neq 2 \text{ and } x \neq 6\}$$

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

Domain: The expression under the square root must be non-negative:
$$x - 1 \geq 0 \implies x \geq 1$$
$$\text{Domain} = [1, +\infty) = \{x \in \mathbf{R} : x \geq 1\}$$

Range: When $x \geq 1$, $x - 1 \geq 0$, so $\sqrt{x-1} \geq 0$. As $x$ increases from $1$ to $\infty$, $f(x)$ increases from $0$ to $\infty$.
$$\text{Range} = [0, +\infty) = \{y \in \mathbf{R} : y \geq 0\}$$

Given: $f(x) = |x - 1|$.

Domain: The absolute value function is defined for all real numbers.
$$\text{Domain} = \mathbf{R}$$

Range: For any $x \in \mathbf{R}$, $|x - 1| \geq 0$. The minimum value is $0$ (at $x = 1$), and the function can take any non-negative value.
$$\text{Range} = [0, +\infty) = \{y \in \mathbf{R} : y \geq 0\}$$

Given: $f(x) = \dfrac{x^2}{1 + x^2}$, $x \in \mathbf{R}$.

Let $y = \dfrac{x^2}{1+x^2}$.

Step 1: Note that $x^2 \geq 0$ and 1 + x^2 \geq 1 > 0, so $y \geq 0$.

Step 2: Also, $y = \dfrac{x^2}{1+x^2} = 1 - \dfrac{1}{1+x^2}$.

Since $1 + x^2 \geq 1$, we have $\dfrac{1}{1+x^2} \leq 1$, so y = 1 - \dfrac{1}{1+x^2} < 1.

Step 3: As $x \to \infty$, $y \to 1$ (but never equals 1). At $x = 0$, $y = 0$.

For any $y_0 \in [0,1)$, we can solve: $y_0 = \dfrac{x^2}{1+x^2} \Rightarrow x^2 = \dfrac{y_0}{1-y_0} \geq 0$, which has a real solution.

\text{Range of } f = [0, 1) = \left\{y \in \mathbf{R} : 0 \leq y < 1\right\}