**2. **Number of elements in set A = 3 and Number of elements in set B = 3

∴ Number of elements in A × B = 3 × 3 = 9

**3. **Given: G = {7, 8} and H = {5, 4, 2}

∴ G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

And H × G = {(5, 7), (4, 7), (2, 7), (5, 8), (4, 8), (2, 8)}

**4. **(i) Here P = {m, n} and Q = {n, m}

Number of elements in set P = 2 and Number of elements in set Q = 2

Number of elements in P × Q = 2 2 = 4

But P × Q = {(m, n), (n, m)} and here number of elements in P × Q = 2

Therefore, statement is false.

Correct statement is P × Q = {(m, m), (n, n), (n, m), (m, n)

(ii) True

(iii) True

**5. ** Here A = {-1, 1}

A × A = {(-1, 1), (-1, 1), (1, -1), (1, 1)}

∴ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, 1, -1), (1, 1, 1)}

**6. ** Given: A × B = {(a, x), (a, y), (b, x), (b, y)}

∴ A = set of first elements = {a, b} and B = set of second elements = {x, y}

**7. **Given: A = {1, 2}, B = {1, 2, 3, 4},

C = {5, 6} and D = {5, 6, 7, 8}

(i) B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

∴ A × B ∩ C = {1, 2} × Φ = Φ ….(i)

A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)

∴ (A × B) ∩ (A × C) = Φ ….(ii)

Therefore, from eq. (i) and (ii), A × B ∩ C

= (A × B) ∩ (A × C)

(ii) A × C = {(1, 5), (1, 6), (2, 5), (2, 6)

B × 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),

Therefore, it is clear that each element of A C is present in B × D.

∴ A × C ⊂ B × D.

**8. **Given: A = {1, 2} and B = {3, 4}

∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Number of elements in A × B = 4

Therefore, Number of subsets of A × B = 2^{4} = 16

Φ,{(2,3)},{(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)},{(1, 3), (1, 4), (2, 3), (2, 4)}

**9. **Here (x, 1) ∈ A × B

⇒ x ∈ A and 1 ∈ B

(y, 2) ∈ A × B

⇒ y ∈ A and 2 ∈ B

(z, 1) ∈ A × B

⇒ z ∈ A and 1 ∈ B

But it is given n(A) = 3 and n(B) = 2

∴ A = {x, y, z} and B = {1, 2}.

**10. **Here (-1, 0) ∈ A × A

⇒ -1 ∈ A and 0 ∈ A

(0, 1) ∈ A × A

⇒ 0 ∈ A and 1 ∈ A

∴ -1, 0, 1 ∈ A

But it is given that n(A × A) = 9 which implies that n(A) = 3

∴ A = {-1, 0, 1}

And A × A = {(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)}

Therefore, the remaining elements of A × A are

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

**11. **Given: A = {1, 2, 3, ……….., 14}

The ordered pairs which satisfy 3x – y = 0 are (1, 3), (2, 6), (3, 9) and (4, 12).

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

Domain = {1, 2, 3, 4}

Range = {3, 6, 9, 12}

Co-domain = {1, 2, 3, ……….., 14}.

**12. ** Given R = {(x, y : y = x + 5, x is a natural number less than 4 : x, y ∈ N)}

Putting x = 1, 2, 3 in y = x + 5 we get = 6, 7, 8

∴ R = {(1, 6), (2, 7), (3, 8)}

Domain = {1, 2, 3}

Range = {6, 7, 8}

**13. **Given: A = {1, 2, 3, 5} and B = {4, 6, 9}, x ∈ A, y ∈ B

∴ x – y = (1 – 4), (1 – 6), (1 – 9), (2 – 4), (2 – 6), (2 – 9), (3 – 4), (3 – 6) (3 – 9),(5 – 4), (5 – 6), (5 – 9)

⇒ x – y = -3, -5, -8, -2, -4, -7, -1, -3, -6, 1, -1, -4

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

**14. **(i) Relation R in set-builder form is R = {(x, y): y = x – 2: x = 5, 6, 7}

(ii) Relation R in roster form is R = {(5, 3), (6, 4), (7, 5)}

Domain = {5, 6, 7}

Range = {3, 4, 5}

**15.** Given: A = {1, 2, 3, 4, 6}

A set of ordered pairs where is exactly divisible by a.

(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}.

**16. ** Given: R = {(x, x + 5): x ∈ (0, 1, 2, 3, 4, 5)} = {(a, b): a = 0, 1, 2, 3, 4, 5}

∴ a = x and b = x + 5

Putting a = 0, 1, 2, 3, 4, 5 we get b = 5, 6, 7, 8, 9, 10

∴ Domain of R = {0, 1, 2, 3, 4, 5}

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

**17. ** Given: R = {(x, x³): x is prime number less than 10}

Putting x = 2, 3, 5, 7

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

**18. **Given: A = and B = {1, 2}

Number of elements in set A = 3 and Number of elements in set B = 2

∴ Number of subsets of A × B = 3 × 2 = 6

Number of relations from A to B = 2^{6}.

**19. ** Given: R = {(a, b): a, b ∈ Z, a – b is an integer}

= {(a, b): a, b ∈ Z, both a and b are even or both a and b are odd}

= {(a, b): a, b ∈ Z, (a and b are even) ∪ (a and b are odd)}

∴ Domain of R = Z

Range of R = Z

**20. **(i) Given Relation is {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

All values of x are distinct. Each value of has a unique value of y.

Therefore, the relation is a function.

∴ Domain of function = {2, 5, 8, 11, 14, 17}

Range of function = {1}

(ii) Given: Relation is {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

All values of x are distinct. Each value of x has a unique value of y.

Therefore, the relation is a function.

∴ Domain of function = {2, 4, 6, 8, 10, 12, 14}

Range of function = {1, 2, 3, 4, 5, 6, 7}

(iii) Given: Relation is {(1, 3), (1, 5), (2, 5)}

This relation is not a function because there is an element 1 which is associated to two elements 3 and 5.

**21. ** (i) Given: f(x) = -|x|. The function is defined for all real values of x.

∴ Domain of the function = R

Now, when x < 0, then |x| = -x

∴ f(x) = -(-x), x < 0

When x = 0, |x| = 0

∴ f(x) = -|0| = 0

When x > 0, |x| = x

∴ f(x) = -x < 0

Therefore, f(x) ≤ 0 for all real values of x.

∴ Range of function = (-∞, 0]

The function is not defined when 9 – x^{2} < 0

∴ Domain of function = {x : 9 – x^{2} ≥ 0} = {x : x^{2} – 9 ≤ 0}

= {x : (x + 3) (x – 3) ≤ 0} = [-3, 3]

∴ Range of function = [0, 3]

**22.** Given: f(x) = 2x – 5

(i) Putting x = 0,

f(0) = 2 × 0 – 5 = -5

(ii) Putting x = 7,

f(7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) Putting x = -3,

f(-3) = 2 × (-3) – 5 = 6 – 5 = -11

**24. ** (i) Given: f(x) = 2 – 3x, x ∈ R and x > 0

∴ 3x > 0 ⇒ -3x < 0

⇒ 2 – 3x < 2

∴ Range of function

= {a ∈ R : a < 2} = (-∞, 2)

(ii) Given: f(x) = x^{2} + 2, x ∈ R

∴ x^{2} ≥ 0 for x ∈ R

⇒ x^{2} + 2 ≥ 2

∴ Range of function

= {a ∈ R : a ≥ 2∀a ∈ R} = [2, ∞)

(iii) Given: f(x) = x, x ∈ R

∴ Range of function = R

**25. ** Given: f(x) = x^{2} 0 ≤ x ≤ 3 and

f(x) = 3x 3 ≤ x ≤ 10

At x = 3, f(3) = (3)^{2} = 9 and

f(3) = 3 × 3 = 9

It is observed that f(x) takes unique value at each point in its domain [0, 10]. Therefore, f is a function.

Now, g(x) = x^{2} 0 ≤ x ≤ 2 and

g(x) = 3x 2 ≤ x ≤ 10

At x = 2, g(2) = (2)^{2} = 4 and

g(2) = 3 × 2 = 6

Therefore, g(x) does not have unique value at x = 2.

Hence, g(x) is not a function.

**26.** Given: f(x) = x^{2}

At x = 1.1 f(1.1) = (1.1)^{2} = 1.21

and f(1) = (1)^{2} = 1

f(x) is a rational function of x.

f(x) assumes real values of all x except for those values of x for which

x^{2} – 8x + 12 = 0

⇒ (x – 6) (x – 2) = 0

⇒ x = 2, 6

∴ Domain of function = R – {2, 6}

**28.** Given: f(x) assumes real values if x – 1 ≥ 0 ⇒ x ≥ 1 ⇒ x ∈ [1, ∞)

∴ Domain of f(x) = [1, ∞)

For x ≥ 1, f(x) ≥ 0

∴ Range of f(x) = all real numbers ≥ 0 = [0, ∞)

**29.** Given: f(x) = |x – 1|

The function f(x) is defined for all values of x.

∴ Domain of f(x) = R

When x > 1, |x – 1| = x – 1 > 0

When x = 1, |x – 1| = 0

When x < 1, |x – 1| = -x + 1 > 0

**31. **Given: f(x) = x + 1 and g(x) = 2x – 3

Now, (f + g) (x) = f(x) + g(x) = x + 1 + 2x – 3 = 3x – 2

And (f – g) (x) = f(x) – g(x) = x + 1 – 2x + 3 = -x + 4

**32.** Given: f(x) = ax + b and

f = {(1, 1), (2, 3), (0, -1), (1, -3)}

⇒ f(1) = 1, f(2) = 3, f(0) = -1, f(-1) = -3

Now f(1) = 1 ⇒ a × 1 + b = 1

⇒ a + b = 1 …(i)

And f(2) = 3 ⇒ a × 2 + b = 3

⇒ 2a + b = 3 …(ii)

Solving eq.(i) and (ii), we get a = 2 and b = -1.

**33. ** Given: R = {(a, b): a, b ∈ N and a = b^{2}}

(i) No, (3, 3) ∉ R because 3 ≠ 3

(ii) No, (9, 3) ∈ R but (3, 9) ∉ R

(iii) No, (81, 9) ∈ R and (9, 3) ∈ R but (81, 3) ∉ R.

**34. **(i) Here A = {1, 2, 3,4} and B = {1, 5, 9, 11, 15, 16}

∴ -A × B = {(1, 1), (1, 5), (1,9), (1, 11), (1, 15), (1, 16) ,(2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

f = {(1, 5),(2, 9), (3, 1), (4, 5), (2, 11)}

Now, (1, 5),(2, 9), (3, 1), (4, 5), (2, 11) ∈ A × B

∴ -f is a relation from A to B.

(ii) f is not a function because (2, 9) ∈ f and (2, 11) ∈ f.

**35. ** We observed that 1 × 4 = 4 and 2 × 2 = 4

⇒ (1 × 4, 1 + 4) ∈ f and (2 × 2, 2 + 2) ∈ f

⇒ (4, 5) ∈ f and (4, 4) ∈ f

It shows that f is not a function from Z to Z.

**36. ** Here A = {9, 10, 11, 12, 13}

For n = 9, f(9) = 3

[∵ 10 = 2 × 5 and 5 is highest prime factor of 10]

For n = 11, f(11) = 11 = 11

[∵ 11 = 1 × 11 and 11 is highest prime factor of 11]

For n = 12, f(12) = 3

[∵ 12 = 3 × 3 × 2 and 3 is highest prime factor of 12]

For n = 13, f(13) = 13

[∵ 13 = 1 × 13 and 13 is highest prime factor of 13]

∴ Range of f = {5, 11, 3, 13}

= {3, 5, 11, 13}