Exercise 2.1 - Question 1

If $\left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)$, find the values of $x$ and $y$.

Solution: Equating the corresponding components of the ordered pairs:

$\frac{x}{3} + 1 = \frac{5}{3} \implies \frac{x}{3} = \frac{5}{3} - 1 = \frac{2}{3} \implies x = 2$

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

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

Exercise 2.1 - Question 2

If the set $A$ has 3 elements and the set $B = \{3, 4, 5\}$, then find the number of elements in $(A \times B)$.

Solution: Given $n(A) = 3$ and $n(B) = 3$. The number of elements in the Cartesian product $A \times B$ is given by $n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9$.

Answer: The number of elements in $A \times B$ is 9.

Exercise 2.1 - Question 3

If $G = \{7, 8\}$ and $H = \{5, 4, 2\}$, find $G \times H$ and $H \times G$.

Solution:

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

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

Exercise 2.1 - Question 4

State whether each statement is true or false. If false, rewrite correctly.

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

Solution: False. $P \times Q = \{(m, n), (m, m), (n, n), (n, m)\}$.

(ii) If $A, B$ are non-empty sets, $A \times B$ is a non-empty set of ordered pairs $(x, y)$ such that $x \in A, y \in B$.

Solution: True.

(iii) If $A = \{1, 2\}, B = \{3, 4\}$, then $A \times (B \cap \phi) = \phi$.

Solution: True, since $B \cap \phi = \phi$ and $A \times \phi = \phi$.

Exercise 2.1 - Question 5

If $A = \{-1, 1\}$, find $A \times A \times A$.

Solution:

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

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

Exercise 2.1 - Question 6

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

Solution: $A$ is the set of first elements of the ordered pairs: $A = \{a, b\}$. $B$ is the set of second elements: $B = \{x, y\}$.

Exercise 2.1 - Question 7

Let $A = \{1, 2\}$, $B = \{1, 2, 3, 4\}$, $C = \{5, 6\}$, $D = \{5, 6, 7, 8\}$. Verify (i) $A \times (B \cap C) = (A \times B) \cap (A \times C)$ (ii) $A \times C \subset B \times D$.

Solution: (i) $B \cap C = \phi$, so $A \times (B \cap C) = \phi$. Also $(A \times B) \cap (A \times C) = \phi$. Hence verified. (ii) $A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$. Since all these pairs belong to $B \times D$, it is a subset.

Exercise 2.1 - Question 8

Let $A = \{1, 2\}$ and $B = \{3, 4\}$. Write $A \times B$. How many subsets will it have? List them.

Solution: $A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$. Number of elements is 4, so subsets = $2^4 = 16$.

Exercise 2.1 - Question 9

Let $n(A) = 3, n(B) = 2$. $(x, 1), (y, 2), (z, 1) \in A \times B$. Find $A$ and $B$.

Solution: $A = \{x, y, z\}$ and $B = \{1, 2\}$.

Exercise 2.1 - Question 10

$A \times A$ has 9 elements, including $(-1, 0)$ and $(0, 1)$. Find $A$ and remaining elements of $A \times A$.

Solution: $A = \{-1, 0, 1\}$. Remaining elements are $\{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1\}$.

Exercise 2.2 - Question 1

$A = \{1, 2, \dots, 14\}$, $R = \{(x, y) : 3x - y = 0\}$. Find domain, codomain, range.

Solution: $R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$. Domain = $\{1, 2, 3, 4\}$, Range = $\{3, 6, 9, 12\}$, Codomain = $A$.

Exercise 2.2 - Question 2

$R = \{(x, y) : y = x + 5, x < 4, x, y \in \mathbb{N}\}$.

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

Exercise 2.2 - Question 3

$A = \{1, 2, 3, 5\}, B = \{4, 6, 9\}$, $R = \{(x, y) : |x - y| \text{ is odd}\}$.

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

Exercise 2.2 - Question 4

Figure relationship $P$ and $Q$.

Solution: (i) $R = \{(x, y) : y = x - 2, x = 5, 6, 7\}$ (ii) $R = \{(5, 3), (6, 4), (7, 5\}$, Domain = $\{5, 6, 7\}$, Range = $\{3, 4, 5\}$.

Exercise 2.2 - Question 5

$A = \{1, 2, 3, 4, 6\}$, $R = \{(a, b) : b \text{ is divisible by } a\}$.

Solution: (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 = $A$ (iii) Range = $A$.

Exercise 2.2 - Question 6

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

Solution: Domain = $\{0, 1, 2, 3, 4, 5\}$, Range = $\{5, 6, 7, 8, 9, 10\}$.

Exercise 2.2 - Question 7

$R = \{(x, x^3) : x \text{ is prime } < 10\}$.

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

Exercise 2.2 - Question 8

$A = \{x, y, z\}, B = \{1, 2\}$. Find number of relations.

Solution: $n(A \times B) = 3 \times 2 = 6$. Total relations = $2^6 = 64$.

Exercise 2.2 - Question 9

$R = \{(a, b) : a - b \text{ is integer}\}$.

Solution: Domain = $\mathbb{Z}$, Range = $\mathbb{Z}$.

Exercise 2.3 - Question 1

Determine which relations are functions, domain, and range.

Solution: (i) Yes, Domain = $\{2, 5, 8, 11, 14, 17\}$, Range = $\{1\}$. (ii) Yes, Domain = $\{2, 4, 6, 8, 10, 12, 14\}$, Range = $\{1, 2, 3, 4, 5, 6, 7\}$. (iii) No, as 1 has two images 3 and 5.

Exercise 2.3 - Question 2

Find domain and range of (i) $f(x) = -|x|$ (ii) $f(x) = \sqrt{9 - x^2}$.

Solution: (i) Domain = $\mathbb{R}$, Range = $(-\infty, 0]$. (ii) Domain = $[-3, 3]$, Range = $[0, 3]$.

Exercise 2.3 - Question 3

$f(x) = 2x - 5$. Find $f(0), f(7), f(-3)$.

Solution: (i) $f(0) = -5$ (ii) $f(7) = 9$ (iii) $f(-3) = -11$.

Exercise 2.3 - Question 4

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

Solution: (i) $t(0) = 32$ (ii) $t(28) = \frac{412}{5}$ (iii) $t(-10) = 14$ (iv) $C = 100$.

Exercise 2.3 - Question 5

Find ranges for given functions.

Solution: (i) $(-\infty, 2)$ (ii) $[2, \infty)$ (iii) $\mathbb{R}$.

Miscellaneous - Question 1

Show $f$ is a function and $g$ is not.

Solution: At $x = 3$, $f(x)$ is uniquely defined. For $g(x)$, at $x = 2$, $g(2)$ gives two values ($4$ and $6$), hence $g$ is not a function.

Miscellaneous - Question 2

If $f(x) = x^2$, find $\frac{f(1.1) - f(1)}{1.1 - 1}$.

Solution: $\frac{1.21 - 1}{0.1} = \frac{0.21}{0.1} = 2.1$.

Miscellaneous - Question 3

Domain of $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$.

Solution: $x^2 - 8x + 12 = (x-6)(x-2) \neq 0 \implies x \neq 2, 6$. Domain = $\mathbb{R} \setminus \{2, 6\}$.

Miscellaneous - Question 4

Domain and range of $f(x) = \sqrt{x - 1}$.

Solution: Domain = $[1, \infty)$, Range = $[0, \infty)$.

Miscellaneous - Question 5

Domain and range of $f(x) = |x - 1|$.

Solution: Domain = $\mathbb{R}$, Range = $[0, \infty)$.

Miscellaneous - Question 6

Range of $f = \left\{(x, \frac{x^2}{1+x^2}) : x \in \mathbb{R}\right\}$.

Solution: Range = $[0, 1)$.

Miscellaneous - Question 7

$f(x) = x + 1, g(x) = 2x - 3$. Find $f+g, f-g, f/g$.

Solution: $(f+g)x = 3x - 2$, $(f-g)x = -x + 4$, $\left(\frac{f}{g}\right)x = \frac{x+1}{2x-3}, x \neq \frac{3}{2}$.

Miscellaneous - Question 8

Find $a, b$ for $f(x) = ax + b$ given pairs.

Solution: $a = 2, b = -1$.

Miscellaneous - Question 9

$R = \{(a, b) : a = b^2\}$. Are statements true?

Solution: (i) No (ii) No (iii) No.

Miscellaneous - Question 10

$A = \{1, 2, 3, 4\}, B = \{1, 5, 9, 11, 15, 16\}$.

Solution: (i) Yes (ii) No, since element 2 has two distinct images (9 and 11).

Miscellaneous - Question 11

Is $f = \{(ab, a+b) : a, b \in \mathbb{Z}\}$ a function?

Solution: No, because different pairs of $(a, b)$ can yield the same first coordinate while having different second coordinates.

Miscellaneous - Question 12

$A = \{9, 10, 11, 12, 13\}$, $f(n) =$ highest prime factor of $n$.

Solution: $f(9)=3, f(10)=5, f(11)=11, f(12)=3, f(13)=13$. Range = $\{3, 5, 11, 13\}$.