Relations and Functions

Two ordered pairs are equal if and only if their corresponding components are equal i.e. $(a, b) = (c, d)$ ⇔ $a = c$ and $b = d$.

(i) $(a + 1, b - 2) = (3, 1)$

Equating the corresponding components,

⇒ $a + 1 = 3$ and $b - 2 = 1$

⇒ $a = 3 - 1$ and $b = 1 + 2$

⇒ $a = 2$ and $b = 3$

Hence, $\bm{a = 2}$ and $\bm{b = 3}$.

(ii) $\left(\frac{a}{3} + 1,\ b - \frac{2}{3}\right) = \left(\frac{5}{3},\ \frac{1}{3}\right)$

Equating the corresponding components,

⇒ $\frac{a}{3} + 1 = \frac{5}{3}$ and $b - \frac{2}{3} = \frac{1}{3}$

⇒ $\frac{a}{3} = \frac{5}{3} - 1$ and $b = \frac{1}{3} + \frac{2}{3}$

⇒ $\frac{a}{3} = \frac{2}{3}$ and $b = \frac{3}{3}$

⇒ $a = 2$ and $b = 1$

Hence, $\bm{a = 2}$ and $\bm{b = 1}$.

(iii) $(2a, a + b) = (6, 2)$

Equating the corresponding components,

⇒ $2a = 6$ and $a + b = 2$

⇒ $a = 3$

Substituting $a = 3$ in $a + b = 2$,

⇒ $3 + b = 2$

⇒ $b = 2 - 3 = -1$

Hence, $\bm{a = 3}$ and $\bm{b = -1}$.

(iv) $(a + b, 3b - 2) = (7, -5)$

Equating the corresponding components,

⇒ $a + b = 7$ and $3b - 2 = -5$

⇒ $3b = -5 + 2$

⇒ $3b = -3$

⇒ $b = -1$

Substituting $b = -1$ in $a + b = 7$,

⇒ $a - 1 = 7$

⇒ $a = 8$

Hence, $\bm{a = 8}$ and $\bm{b = -1}$.

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

(i) (4x + 3, y) = (3x + 5, −2)

Equating the corresponding components,

⇒ 4x + 3 = 3x + 5 and y = −2

⇒ 4x − 3x = 5 − 3

⇒ x = 2

Hence, x = 2 and y = −2.

(ii) (x − y, x + y) = (6, 10)

Equating the corresponding components,

x − y = 6 ...........(1)

x + y = 10 ...........(2)

Adding equations (1) and (2), we get :

⇒ 2x = 16

⇒ x = 8

Substituting x = 8 in equation (2), we get :

⇒ 8 + y = 10

⇒ y = 2

Hence, x = 8 and y = 2.

Given,

The ordered pairs (a, −1) and (5, b) belong to the set {(x, y) : y = 2x − 3}, so each of them satisfies y = 2x − 3.

For the ordered pair (a, −1), we have x = a and y = −1.

⇒ −1 = 2a − 3

⇒ 2a = −1 + 3

⇒ 2a = 2

⇒ a = 1

For the ordered pair (5, b), we have x = 5 and y = b.

⇒ b = 2(5) − 3

⇒ b = 10 − 3

⇒ b = 7

Hence, a = 1 and b = 7.

Given,

A = {−1, 0, 1} and B = {3, 5}

We know that,

A × B = {(a, b) : for all a ∈ A, b ∈ B}

(i) A × B

Pairing each element of A with each element of B,

Hence, A × B = {(−1, 3), (−1, 5), (0, 3), (0, 5), (1, 3), (1, 5)}.

(ii) B × A

Pairing each element of B with each element of A,

Hence, B × A = {(3, −1), (3, 0), (3, 1), (5, −1), (5, 0), (5, 1)}.

(iii) B × B

Pairing each element of B with each element of B,

Hence, B × B = {(3, 3), (3, 5), (5, 3), (5, 5)}.

We know that,

If A and B are finite sets, then n(A × B) = n(A) × n(B).

Given,

n(A) = 3 and B = {3, 4, 5}, so n(B) = 3.

∴ n(A × B) = n(A) × n(B) = 3 × 3 = 9

Hence, the number of elements in A × B is 9.

Given,

A = {−3, −1, 0, 4} and B = {−1, 0, 1, 2, 3}

∴ n(A) = 4 and n(B) = 5

We know that, n(X × Y) = n(X) × n(Y).

(i) A × B

n(A × B) = n(A) × n(B) = 4 × 5 = 20

Hence, the number of elements in A × B is 20.

(ii) B × A

n(B × A) = n(B) × n(A) = 5 × 4 = 20

Hence, the number of elements in B × A is 20.

(iii) A × A

n(A × A) = n(A) × n(A) = 4 × 4 = 16

Hence, the number of elements in A × A is 16.

(iv) B × B

n(B × B) = n(B) × n(B) = 5 × 5 = 25

Hence, the number of elements in B × B is 25.

Given,

A = {1, 2} and B = {3, 4}

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

⇒ n(A × B) = 4

We know that,

A set containing n elements has 2ⁿ subsets.

∴ Number of subsets of A × B = 2⁴ = 16

Hence, A × B will have 16 subsets.

Given,

x ∈ {2, 3, 5} and y ∈ {2, 4, 6}

Taking each value of x in turn and choosing those values of y which are greater than x,

when x = 2, y = 4, 6 (as 2 < 4 and 2 < 6),

when x = 3, y = 4, 6 (as 3 < 4 and 3 < 6),

when x = 5, y = 6 (as 5 < 6).

Thus, the corresponding ordered pairs are (2, 4), (2, 6), (3, 4), (3, 6) and (5, 6).

Hence, the required set of ordered pairs is {(2, 4), (2, 6), (3, 4), (3, 6), (5, 6)}.

Given,

x ∈ {−1, 2, 3, 4, 5} and y ∈ {0, 3, 6} with x + y = 5 i.e. y = 5 − x.

Taking each value of x in turn,

when x = −1, y = 5 − (−1) = 6 ∈ {0, 3, 6},

when x = 2, y = 5 − 2 = 3 ∈ {0, 3, 6},

when x = 3, y = 5 − 3 = 2 ∉ {0, 3, 6},

when x = 4, y = 5 − 4 = 1 ∉ {0, 3, 6},

when x = 5, y = 5 − 5 = 0 ∈ {0, 3, 6}.

Thus, the corresponding ordered pairs are (−1, 6), (2, 3) and (5, 0).

Hence, the required set of ordered pairs is {(−1, 6), (2, 3), (5, 0)}.

Given,

x ∈ {2, 3, 4} and y ∈ {4, 6, 9, 10}

Taking each value of x in turn and choosing those values of y of which x is a factor,

when x = 2, y = 4, 6, 10 (as 2 divides 4, 6 and 10),

when x = 3, y = 6, 9 (as 3 divides 6 and 9),

when x = 4, y = 4 (as 4 divides 4).

Thus, the corresponding ordered pairs are (2, 4), (2, 6), (2, 10), (3, 6), (3, 9) and (4, 4).

Hence, the required set of ordered pairs is {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4)}.

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

Given,

P = {m, n} and Q = {n, m}

Pairing each element of P with each element of Q,

P × Q = {(m, n), (m, m), (n, n), (n, m)}

which has four ordered pairs and not two.

Hence, the given statement is false and the correct statement is P × Q = {(m, n), (m, m), (n, n), (n, m)}.

(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 ∈ B and y ∈ A

By definition, A × B = {(x, y) : for all x ∈ A, y ∈ B}, so the first component is taken from A and the second component from B.

Hence, the given statement is false and the correct statement is: 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.

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

Given,

A = {1, 2} and B = {3, 4}

The empty set φ has no element, so B and φ have no element in common.

⇒ B ∩ φ = φ

∴ A × (B ∩ φ) = A × φ = φ, as there is no second component available to form an ordered pair.

Hence, the given statement is true.

Given,

A = {−1, 0, 1, 2, 3}

S is the set of those ordered pairs of A × A whose second component is 0, so the first component can be any element of A while the second component is fixed as 0.

Thus, the corresponding ordered pairs are (−1, 0), (0, 0), (1, 0), (2, 0) and (3, 0).

Hence, S = {(−1, 0), (0, 0), (1, 0), (2, 0), (3, 0)}.

Given,

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

We know that,

A is the set of all first components of the ordered pairs of A × B and B is the set of all second components.

The first components are −1, −1, 2, 2, 3, 3.

∴ A = {−1, 2, 3}

The second components are 1, 2, 1, 2, 1, 2.

∴ B = {1, 2}

Hence, A = {−1, 2, 3} and B = {1, 2}.

Given,

$A = {x, y, z}$, so $n(A) = 3$.

Also, $n(A \times B) = 6$

We know that,

$n(A \times B) = n(A) \times n(B)$.

⇒ $6 = 3 \times n(B)$

⇒ $n(B) = \frac{6}{3} = 2$

The ordered pairs $(x, 1)$, $(y, 2)$ and $(z, 1)$ belong to $A \times B$, so their second components 1 and 2 belong to $B$.

As $B$ contains only 2 elements, $B$ can have no element besides 1 and 2.

Hence, $\bm{B = {1, 2}}$.

Given,

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

B × A is obtained from A × B by interchanging the components of each ordered pair.

Thus, the corresponding ordered pairs are (1, x), (2, y), (3, x), (3, y), (1, y) and (2, x).

Hence, B × A = {(1, x), (2, y), (3, x), (3, y), (1, y), (2, x)}.

Given,

A = {1, 2, 3, 4} and B = {5, 7, 9}

(i) Pairing each element of A with each element of B,

Hence, A × B = {(1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)}.

(ii) Pairing each element of B with each element of A,

Hence, B × A = {(5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4), (9, 1), (9, 2), (9, 3), (9, 4)}.

(iii) The ordered pair (1, 5) ∈ A × B, but (1, 5) ∉ B × A, since the pair (1, 5) is not equal to the pair (5, 1).

Hence, A × B ≠ B × A.

(iv) Here n(A) = 4 and n(B) = 3.

∴ n(A × B) = n(A) × n(B) = 4 × 3 = 12

and n(B × A) = n(B) × n(A) = 3 × 4 = 12

Hence, n(A × B) = n(B × A).

Given,

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

The sets B and C have no element in common.

⇒ B ∩ C = φ

∴ A × (B ∩ C) = A × φ

⇒ A × (B ∩ C) = φ ...........(1)

Now,

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

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

These two sets have no ordered pair in common.

∴ (A × B) ∩ (A × C) = φ ...........(2)

From equations (1) and (2), we get

Hence, A × (B ∩ C) = (A × B) ∩ (A × C).

(ii) A × C is a subset of B × D

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)}

Each of the ordered pairs (1, 5), (1, 6), (2, 5) and (2, 6) of A × C is also an element of B × D.

Hence, A × C is a subset of B × D.

Given,

A = {x : x ∈ W, x < 2}

The whole numbers less than 2 are 0 and 1.

∴ A = {0, 1}

B = {x : x ∈ N, 1 < x < 5}

The natural numbers lying between 1 and 5 are 2, 3 and 4.

∴ B = {2, 3, 4}

and C = {3, 5}

(i) A × (B ∩ C)

The only element common to B and C is 3.

⇒ B ∩ C = {3}

∴ A × (B ∩ C) = {0, 1} × {3}

Hence, A × (B ∩ C) = {(0, 3), (1, 3)}.

(ii) A × (B ∪ C)

Collecting all the elements which belong to B or C or both,

⇒ B ∪ C = {2, 3, 4, 5}

∴ A × (B ∪ C) = {0, 1} × {2, 3, 4, 5}

= {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

Hence, A × (B ∪ C) = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}.

Given,

A = {x : x ∈ N and x ≤ 3}

The natural numbers not exceeding 3 are 1, 2 and 3.

∴ A = {1, 2, 3}

B = {x : x ∈ I, −1 ≤ x ≤ 1}

The integers from −1 to 1 are −1, 0 and 1.

∴ B = {−1, 0, 1}

and C = {1, 2}

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

B ∪ C = {−1, 0, 1, 2}

∴ A × (B ∪ C) = {(1, −1), (1, 0), (1, 1), (1, 2), (2, −1), (2, 0), (2, 1), (2, 2), (3, −1), (3, 0), (3, 1), (3, 2)} ...........(1)

Now,

A × B = {(1, −1), (1, 0), (1, 1), (2, −1), (2, 0), (2, 1), (3, −1), (3, 0), (3, 1)} and

A × C = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}

Collecting all the ordered pairs which belong to A × B or A × C or both,

∴ (A × B) ∪ (A × C) = {(1, −1), (1, 0), (1, 1), (1, 2), (2, −1), (2, 0), (2, 1), (2, 2), (3, −1), (3, 0), (3, 1), (3, 2)} ...........(2)

From equations (1) and (2), we get

Hence, A × (B ∪ C) = (A × B) ∪ (A × C).

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

The only element common to B and C is 1.

⇒ B ∩ C = {1}

∴ A × (B ∩ C) = {(1, 1), (2, 1), (3, 1)} ...........(1)

The ordered pairs common to A × B and A × C are (1, 1), (2, 1) and (3, 1).

∴ (A × B) ∩ (A × C) = {(1, 1), (2, 1), (3, 1)} ...........(2)

From equations (1) and (2), we get

Hence, A × (B ∩ C) = (A × B) ∩ (A × C).

(iii) (A − B) × C = A × C − B × C

Removing from A the element 1 which also belongs to B,

⇒ A − B = {2, 3}

∴ (A − B) × C = {(2, 1), (2, 2), (3, 1), (3, 2)} ...........(1)

Now,

A × C = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)} and

B × C = {(−1, 1), (−1, 2), (0, 1), (0, 2), (1, 1), (1, 2)}

Removing from A × C the ordered pairs (1, 1) and (1, 2) which also belong to B × C,

∴ A × C − B × C = {(2, 1), (2, 2), (3, 1), (3, 2)} ...........(2)

From equations (1) and (2), we get

Hence, (A − B) × C = A × C − B × C.

Given,

P = {1, 2}

We know that,

P × P × P = {(a, b, c) : for all a ∈ P, b ∈ P, c ∈ P}

Forming all the triplets whose components are taken from P,

Hence, P × P × 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,

A = {a, b, c}, so n(A) = 3.

Also, n(A × B) = 6

We know that,

n(A × B) = n(A) × n(B).

⇒ 6 = 3 × n(B)

⇒ n(B) = 2

The ordered pairs (a, p), (b, q) and (c, p) belong to A × B, so their second components p and q belong to B.

As n(B) = 2, B can have no element besides p and q.

∴ B = {p, q}

Now,

A × B = {(a, p), (a, q), (b, p), (b, q), (c, p), (c, q)}

Removing the three given ordered pairs (a, p), (b, q) and (c, p),

Hence, B = {p, q} and the remaining ordered pairs of A × B are (a, q), (b, p) and (c, q).

Given,

B = {2, 3, 5}, so n(B) = 3.

The ordered pairs (a, 2), (b, 3) and (c, 5) belong to A × B, so their first components a, b and c belong to A.

For A × B to have the least number of elements, A must have the least number of elements, i.e. A can have no element besides a, b and c.

∴ A = {a, b, c}, thus n(A) = 3

n(A × B) = n(A) × n(B) = 3 × 3 = 9

Now,

A × B = {(a, 2), (a, 3), (a, 5), (b, 2), (b, 3), (b, 5), (c, 2), (c, 3), (c, 5)}

Removing the three given ordered pairs (a, 2), (b, 3) and (c, 5),

Hence, A = {a, b, c} and the remaining ordered pairs of A × B are (a, 3), (a, 5), (b, 2), (b, 5), (c, 2) and (c, 3).

We know that,

A relation from a set X to a set Y is a subset of X × Y, so the number of relations from X to Y is the number of subsets of X × Y i.e. 2^{n(X) × n(Y)}.

Given,

n(A) = 2 and n(B) = 3

(i) n(A × B) = n(A) × n(B) = 2 × 3 = 6

∴ Number of relations from A to B = 2⁶ = 64

Hence, the number of relations from A to B is 64.

(ii) n(B × A) = n(B) × n(A) = 3 × 2 = 6

∴ Number of relations from B to A = 2⁶ = 64

Hence, the number of relations from B to A is 64.

(iii) n(A × A) = n(A) × n(A) = 2 × 2 = 4

∴ Number of relations from A to A = 2⁴ = 16

Hence, the number of relations from A to A is 16.

Given,

A = {1, 2} and B = {3, 4}

(i) Pairing each element of A with each element of B,

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

(ii) A relation from A to B is a subset of A × B, so the number of relations from A to B is the number of subsets of A × B.

From part (i), n(A × B) = 4.

∴ Number of relations from A to B = 2⁴ = 16

Hence, the number of relations from A to B is 16.

Given,

R = {(−2, 1), (0, 2), (3, 1), (0, −1), (4, 2), (5, 1)}

We know that,

The domain of a relation is the set of all first components of its ordered pairs and the range is the set of all second components.

The first components are −2, 0, 3, 0, 4, 5.

∴ Domain of R = {−2, 0, 3, 4, 5}

The second components are 1, 2, 1, −1, 2, 1.

∴ Range of R = {1, 2, −1}

Hence, domain of R = {−2, 0, 3, 4, 5} and range of R = {1, 2, −1}.

Given,

A = {2, 3, 5}, B = {2, 4, 6} and R = {(x, y) : x ∈ A, y ∈ B and x < y}

Taking each element of A in turn and choosing those elements of B which are greater than it,

when x = 2, y = 4, 6,

when x = 3, y = 4, 6,

when x = 5, y = 6.

Thus, the corresponding ordered pairs are (2, 4), (2, 6), (3, 4), (3, 6) and (5, 6).

Hence, R = {(2, 4), (2, 6), (3, 4), (3, 6), (5, 6)}.

Given,

A = {1, 3, 5, 7, 8} and B = {2, 3, 4, 6, 8, 10}

R is the relation 'is one less than' from A to B i.e. R = {(x, y) : x ∈ A, y ∈ B and y = x + 1}.

Taking each element of A in turn,

when x = 1, y = 2 ∈ B,

when x = 3, y = 4 ∈ B,

when x = 5, y = 6 ∈ B,

when x = 7, y = 8 ∈ B,

when x = 8, y = 9 ∉ B.

Thus, the corresponding ordered pairs are (1, 2), (3, 4), (5, 6) and (7, 8).

Hence, R = {(1, 2), (3, 4), (5, 6), (7, 8)}.

Given,

A = {2, 3, 4}, B = {4, 6, 9, 10} and R = {(x, y) : (x, y) ∈ A × B such that x is a factor of y}

Taking each element of A in turn and selecting those elements of B that are divisible by it,

when x = 2, y = 4, 6, 10,

when x = 3, y = 6, 9,

when x = 4, y = 4.

Thus, the corresponding ordered pairs are (2, 4), (2, 6), (2, 10), (3, 6), (3, 9) and (4, 4).

Hence, R = {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4)}.

Given,

A = {1, 2, 3, ......, 17}

R = {(x, y) : 3x − y = 0, x, y ∈ A}

⇒ 3x − y = 0

⇒ y = 3x

Taking those values of x ∈ A for which y = 3x also belongs to A,

when x = 1, y = 3 ∈ A,

when x = 2, y = 6 ∈ A,

when x = 3, y = 9 ∈ A,

when x = 4, y = 12 ∈ A,

when x = 5, y = 15 ∈ A,

when x = 6, y = 18 ∉ A.

For all the remaining values of x ∈ A, y = 3x is greater than 17 and so does not belong to A.

Thus, the corresponding ordered pairs are (1, 3), (2, 6), (3, 9), (4, 12) and (5, 15).

Hence, R = {(1, 3), (2, 6), (3, 9), (4, 12), (5, 15)}.

The prime numbers less than 10 are 2, 3, 5 and 7.

(i) R = {(x, x³) : x is a prime number less than 10}

Taking each prime number in turn,

when x = 2, x³ = 8,

when x = 3, x³ = 27,

when x = 5, x³ = 125,

when x = 7, x³ = 343.

Thus, the corresponding ordered pairs are (2, 8), (3, 27), (5, 125) and (7, 343).

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

(ii) R = {(x − 2, x²) : x is a prime number less than 10}

Taking each prime number in turn,

when x = 2, x − 2 = 0 and x² = 4,

when x = 3, x − 2 = 1 and x² = 9,

when x = 5, x − 2 = 3 and x² = 25,

when x = 7, x − 2 = 5 and x² = 49.

Thus, the corresponding ordered pairs are (0, 4), (1, 9), (3, 25) and (5, 49).

Hence, R = {(0, 4), (1, 9), (3, 25), (5, 49)}.

Given,

R = {(1, −1), (2, 0), (3, 1), (4, 2), (5, 3)}

(i) The first components of the ordered pairs of R are 1, 2, 3, 4, 5 and the second components are −1, 0, 1, 2, 3.

Hence, domain of R = {1, 2, 3, 4, 5} and range of R = {−1, 0, 1, 2, 3}.

(ii) In each ordered pair of R, the second component is 2 less than the first component i.e.

−1 = 1 − 2, 0 = 2 − 2, 1 = 3 − 2, 2 = 4 − 2, 3 = 5 − 2

So each ordered pair is of the form (x, x − 2), where x is a natural number from 1 to 5.

Hence, R = {(x, y) : x ∈ N, 1 ≤ x ≤ 5, y = x − 2}.

Given,

A = {2, 4, 6} and B = {4, 6, 18}

R is the relation 'is a factor of' from A to B i.e. R = {(x, y) : x ∈ A, y ∈ B and x is a factor of y}.

Taking each element of A in turn and choosing those elements of B of which it is a factor,

when x = 2, y = 4, 6, 18,

when x = 4, y = 4,

when x = 6, y = 6, 18.

Thus, the corresponding ordered pairs are (2, 4), (2, 6), (2, 18), (4, 4), (6, 6) and (6, 18).

∴ R = {(2, 4), (2, 6), (2, 18), (4, 4), (6, 6), (6, 18)}

Representing R by an arrow diagram, we draw the elements of A in one closed figure and those of B in another, and join x to y by an arrow whenever (x, y) ∈ R.

Hence, R = {(2, 4), (2, 6), (2, 18), (4, 4), (6, 6), (6, 18)} and R is represented by the arrow diagram given above.

Given,

R₁ = {(x, | x |) : x is a real number}

The first component x can be any real number, since | x | is defined for every real number x.

∴ Domain of R₁ = R

The second component is | x |, and we know that | x | ≥ 0 for every real number x. Also every non-negative real number k is the modulus of the real number k itself.

∴ Range of R₁ = [0, ∞)

Hence, domain of R₁ = R and range of R₁ = [0, ∞).

Given,

R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}

Taking each value of x in turn,

when x = 0, x + 5 = 5,

when x = 1, x + 5 = 6,

when x = 2, x + 5 = 7,

when x = 3, x + 5 = 8,

when x = 4, x + 5 = 9,

when x = 5, x + 5 = 10.

Thus, the corresponding ordered pairs are (0, 5), (1, 6), (2, 7), (3, 8), (4, 9) and (5, 10).

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

Hence, domain of R = {0, 1, 2, 3, 4, 5} and range of R = {5, 6, 7, 8, 9, 10}.

From the given arrow diagram,

P = {5, 6, 7} and Q = {3, 4, 5}

and the arrows join 5 to 3, 6 to 4 and 7 to 5.

(i) Writing the ordered pairs joined by the arrows,

Thus, the corresponding ordered pairs are (5, 3), (6, 4) and (7, 5).

Hence, the relation in roster form is {(5, 3), (6, 4), (7, 5)}.

(ii) In each ordered pair the second component is 2 less than the first component i.e.

3 = 5 − 2, 4 = 6 − 2, 5 = 7 − 2

So each ordered pair is of the form (x, y) where y = x − 2 and x is a natural number from 5 to 7.

Hence, the relation in set builder form is {(x, y) : y = x − 2, x ∈ N, 5 ≤ x ≤ 7}.

The first components of the ordered pairs are 5, 6, 7 and the second components are 3, 4, 5.

Hence, domain of the relation = {5, 6, 7} and range of the relation = {3, 4, 5}.

Given,

R = {(a, b) : a ∈ N, b ∈ N and a + 3b = 12}

a + 3b = 12 ⇒ a = 12 − 3b

(i) Taking those values of b ∈ N for which a = 12 − 3b is also a natural number,

when b = 1, a = 12 − 3 = 9 ∈ N,

when b = 2, a = 12 − 6 = 6 ∈ N,

when b = 3, a = 12 − 9 = 3 ∈ N,

when b = 4, a = 12 − 12 = 0 ∉ N.

For all the values of b greater than 4, a = 12 − 3b is negative and so does not belong to N.

Thus, the corresponding ordered pairs are (9, 1), (6, 2) and (3, 3).

Hence, R = {(9, 1), (6, 2), (3, 3)}.

(ii) The first components of the ordered pairs of R are 9, 6 and 3.

Hence, domain of R = {9, 6, 3}.

(iii) The second components of the ordered pairs of R are 1, 2 and 3.

Hence, range of R = {1, 2, 3}.

Given,

R = {(x, y) : x, y ∈ W, x² + y² = 100}

As x and y are whole numbers and x² + y² = 100, neither x² nor y² can exceed 100, so x and y can take the values 0, 1, 2, ....., 10 only.

Taking each value of x in turn,

when x = 0,

⇒ 0² + y² = 100

⇒ y² = 100

⇒ y = 10 ∈ W

when x = 6,

⇒ 6² + y² = 100

⇒ y² = 100 − 36 = 64

⇒ y = 8 ∈ W

when x = 8,

⇒ 8² + y² = 100

⇒ y² = 100 − 64 = 36

⇒ y = 6 ∈ W

when x = 10,

⇒ 10² + y² = 100

⇒ y² = 100 − 100 = 0

⇒ y = 0 ∈ W

For all the other values of x, y² = 100 − x² is not the square of a whole number.

Thus, the corresponding ordered pairs are (0, 10), (6, 8), (8, 6) and (10, 0).

∴ R = {(0, 10), (6, 8), (8, 6), (10, 0)}

The first components are 0, 6, 8, 10 and the second components are 10, 8, 6, 0.

Hence, domain of R = {0, 6, 8, 10}, range of R = {0, 6, 8, 10} and R = {(0, 10), (6, 8), (8, 6), (10, 0)}.

Given,

$R = \bigg\lbrace(x, y) : y = x + \frac{12}{x},\ x, y \in \textbf{N}\bigg\rbrace$

As $x \in \textbf{N}$ and $y = x + \frac{12}{x}$ is a natural number, $\frac{12}{x}$ must be a natural number, so $x$ must be a positive divisor of 12.

∴ $x = 1, 2, 3, 4, 6, 12$

(i) Taking each value of $x$ in turn,

when $x = 1$,

⇒ $y = 1 + \frac{12}{1}$

⇒ $y = 1 + 12$

⇒ $y = 13$

when $x = 2$,

⇒ $y = 2 + \frac{12}{2}$

⇒ $y = 2 + 6$

⇒ $y = 8$

when $x = 3$,

⇒ $y = 3 + \frac{12}{3}$

⇒ $y = 3 + 4$

⇒ $y = 7$

when $x = 4$,

⇒ $y = 4 + \frac{12}{4}$

⇒ $y = 4 + 3$

⇒ $y = 7$

when $x = 6$,

⇒ $y = 6 + \frac{12}{6}$

⇒ $y = 6 + 2$

⇒ $y = 8$

when $x = 12$,

⇒ $y = 12 + \frac{12}{12}$

⇒ $y = 12 + 1$

⇒ $y = 13$

Thus, the corresponding ordered pairs are $(1, 13)$, $(2, 8)$, $(3, 7)$, $(4, 7)$, $(6, 8)$ and $(12, 13)$.

Hence, $\bm{R = {(1, 13), (2, 8), (3, 7), (4, 7), (6, 8), (12, 13)}}$.

(ii) The first components of the ordered pairs of $R$ are 1, 2, 3, 4, 6 and 12.

Hence, domain of $\bm{R = {1, 2, 3, 4, 6, 12}}$.

(iii) The second components of the ordered pairs of $R$ are 13, 8, 7, 7, 8 and 13.

Hence, range of $\bm{R = {13, 8, 7}}$.

Given,

The relation is {(x, y) : x = 3y, x ∈ N, y ∈ N and x, y < 10}

Taking those values of y less than 10 for which x = 3y is also less than 10,

when y = 1, x = 3, which is less than 10,

when y = 2, x = 6 which is less than 10,

when y = 3, x = 9 which is less than 10,

when y = 4, x = 12 which is not less than 10.

For all the values of y greater than 4 also, x = 3y is not less than 10.

Thus, the corresponding ordered pairs are (3, 1), (6, 2) and (9, 3).

∴ The relation = {(3, 1), (6, 2), (9, 3)}

The first components are 3, 6, 9 and the second components are 1, 2, 3.

Hence, domain of the relation = {3, 6, 9} and range of the relation = {1, 2, 3}.

Given,

A = {−2, −1, 0, 1, 2}

(i) Here R = {(x, y) : x, y ∈ A and x > y}.

Taking each element of A in turn as the first component and choosing those elements of A which are less than it,

when x = −1, y = −2,

when x = 0, y = −2, −1,

when x = 1, y = −2, −1, 0,

when x = 2, y = −2, −1, 0, 1.

For x = −2, there is no element of A less than −2.

Thus, the corresponding ordered pairs are (−1, −2), (0, −2), (0, −1), (1, −2), (1, −1), (1, 0), (2, −2), (2, −1), (2, 0) and (2, 1).

Hence, R = {(−1, −2), (0, −2), (0, −1), (1, −2), (1, −1), (1, 0), (2, −2), (2, −1), (2, 0), (2, 1)}.

(ii) Here R = {(x, y) : x, y ∈ A and x = y²}.

Taking each element of A in turn as the second component,

when y = −2, y² = 4 ∉ A,

when y = −1, y² = 1 ∈ A,

when y = 0, y² = 0 ∈ A,

when y = 1, y² = 1 ∈ A,

when y = 2, y² = 4 ∉ A.

Thus, the corresponding ordered pairs are (1, −1), (0, 0) and (1, 1).

Hence, R = {(0, 0), (1, −1), (1, 1)}.

(iii) Here R = {(x, y) : x, y ∈ A and x = −y}.

Taking each element of A in turn as the second component,

when y = −2, x = 2 ∈ A,

when y = −1, x = 1 ∈ A,

when y = 0, x = 0 ∈ A,

when y = 1, x = −1 ∈ A,

when y = 2, x = −2 ∈ A.

Thus, the corresponding ordered pairs are (2, −2), (1, −1), (0, 0), (−1, 1) and (−2, 2).

Hence, R = {(2, −2), (1, −1), (0, 0), (−1, 1), (−2, 2)}.

Given,

A = {1, 3, 5, 6} and B = {3, 4, 5}

(i) R = {(x, y) : (x, y) ∈ A × B : x + y is even}

The sum x + y is even only when x and y are both odd or both even.

The odd elements of A are 1, 3, 5 and the odd elements of B are 3, 5.

⇒ The pairs with both components odd are (1, 3), (1, 5), (3, 3), (3, 5), (5, 3), (5, 5).

The only even element of A is 6 and the only even element of B is 4.

⇒ The pair with both components even is (6, 4).

Thus, the corresponding ordered pairs are (1, 3), (1, 5), (3, 3), (3, 5), (5, 3), (5, 5) and (6, 4).

Hence, R = {(1, 3), (1, 5), (3, 3), (3, 5), (5, 3), (5, 5), (6, 4)}.

(ii) R = {(x, y) : (x, y) ∈ A × B : xy is odd}

The product xy is odd only when x and y are both odd.

The odd elements of A are 1, 3, 5 and the odd elements of B are 3, 5.

Thus, the corresponding ordered pairs are (1, 3), (1, 5), (3, 3), (3, 5), (5, 3) and (5, 5).

Hence, R = {(1, 3), (1, 5), (3, 3), (3, 5), (5, 3), (5, 5)}.

Given,

R = {(x, y) : x, y ∈ Z, y = 2x − 4}

As (a, −2) ∈ R, the pair (a, −2) satisfies y = 2x − 4.

⇒ −2 = 2a − 4

⇒ 2a = −2 + 4

⇒ 2a = 2

⇒ a = 1

As (4, b²) ∈ R, the pair (4, b²) satisfies y = 2x − 4.

⇒ b² = 2(4) − 4

⇒ b² = 8 − 4

⇒ b² = 4

⇒ b = ±2

Hence, a = 1 and b = ±2.

Let the linear relation between the components x and y be y = mx + c.

(i) R = {(−1, −1), (0, 2), (1, 5), ......}

As (0, 2) ∈ R,

⇒ 2 = m(0) + c

⇒ c = 2

As (1, 5) ∈ R,

⇒ 5 = m(1) + 2

⇒ m = 5 − 2 = 3

∴ y = 3x + 2

Checking with the pair (−1, −1), 3(−1) + 2 = −3 + 2 = −1, which is the second component.

Hence, the required linear relation is y = 3x + 2.

(ii) R = {(0, 2), (−1, 5), (2, −4), ......}

As (0, 2) ∈ R,

⇒ 2 = m(0) + c

⇒ c = 2

As (−1, 5) ∈ R,

⇒ 5 = m(−1) + 2

⇒ −m = 5 − 2

⇒ m = −3

∴ y = −3x + 2 = 2 − 3x

Checking with the pair (2, −4), 2 − 3(2) = 2 − 6 = −4, which is the second component.

Hence, the required linear relation is y = 2 − 3x.

We know that,

A relation is a function if and only if no two different ordered pairs of the relation have the same first component.

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

The ordered pairs (1, 1) and (1, 2) have the same first component 1, so the element 1 has more than one image.

Hence, the given relation is not a function.

(ii) {(a, b), (b, c), (c, d), (d, e)}

The first components a, b, c, d are all different, so no element has more than one image.

The first components are a, b, c, d and the second components are b, c, d, e.

Hence, the given relation is a function whose domain = {a, b, c, d} and range = {b, c, d, e}.

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

The ordered pairs (1, 2) and (1, 3) have the same first component 1, so the element 1 has two different images 2 and 3.

Hence, the given relation is not a function.

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

The first components 1, 2, 3, 4 are all different, so no element has more than one image.

The first components are 1, 2, 3, 4 and the second component in each pair is 2.

Hence, the given relation is a function whose domain = {1, 2, 3, 4} and range = {2}.

(v) {(2, 1), (0, −1), (3, 1), (5, 4), (−1, 0), (3, 4), (1, 0)}

The ordered pairs (3, 1) and (3, 4) have the same first component 3, so the element 3 has two different images 1 and 4.

Hence, the given relation is not a function.

(vi) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 3), (14, 7)}

The first components 2, 4, 6, 8, 10, 14 are all different, so no element has more than one image.

The first components are 2, 4, 6, 8, 10, 14 and the second components are 1, 2, 3, 4, 3, 7.

Hence, the given relation is a function whose domain = {2, 4, 6, 8, 10, 14} and range = {1, 2, 3, 4, 7}.

Given,

X = {−4, 1, 2, 3} and Y = {a, b, c}

We know that,

A relation from X to Y is a function from X to Y if and only if every element of X has one and only one image in Y.

(i) {(−4, a), (1, a), (2, b)}

The element 3 of X does not occur as the first component of any ordered pair, so 3 has no image.

Hence, the given relation is not a function from X to Y.

(ii) {(−4, b), (1, b), (2, a), (3, c)}

Each of the four elements −4, 1, 2 and 3 of X occurs exactly once as a first component, so every element of X has a unique image in Y.

Hence, the given relation is a function from X to Y.

(iii) {(−4, a), (1, a), (2, b), (3, c), (1, b)}

The ordered pairs (1, a) and (1, b) have the same first component 1, so the element 1 has two different images a and b.

Hence, the given relation is not a function from X to Y.

Given,

A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) In each ordered pair of f, the first component 1, 2, 3, 4 or 2 belongs to A and the second component 5, 9, 1, 5 or 11 belongs to B, so every ordered pair of f is an element of A × B i.e. f ⊂ A × B.

Hence, the given statement is true.

(ii) The ordered pairs (2, 9) and (2, 11) have the same first component 2, so the element 2 of A has two different images 9 and 11 in B.

∴ f is not a function from A to B.

Hence, the given statement is not true.

We know that,

A relation is a function if and only if every element of its domain has one and only one image.

(i) f = {(x, x) : x ∈ R}

Every real number x occurs exactly once as a first component and its image is the unique real number x itself.

Hence, f is a function.

(ii) g = {(n, n²) : n is a positive integer}

Every positive integer n occurs exactly once as a first component and its image is the unique number n², since a number has only one square.

Hence, g is a function.

(iii) h = {(x, 3) : x ∈ R}

Every real number x occurs exactly once as a first component and its image is the unique number 3. Different elements of the domain may have the same image, which is allowed in a function.

Hence, h is a function.

(i) Substituting each value of x in f(x) = x² + 1,

⇒ f(−1) = (−1)² + 1 = 1 + 1 = 2,

⇒ f(0) = 0² + 1 = 0 + 1 = 1,

⇒ f(1) = 1² + 1 = 1 + 1 = 2,

⇒ f(3) = 3² + 1 = 9 + 1 = 10.

Thus, the corresponding ordered pairs are (−1, 2), (0, 1), (1, 2) and (3, 10).

Hence, f = {(−1, 2), (0, 1), (1, 2), (3, 10)}.

(ii) Substituting each value of x in f(x) = 2x − 1,

⇒ f(−2) = 2(−2) − 1 = −4 − 1 = −5,

⇒ f(0) = 2(0) − 1 = 0 − 1 = −1,

⇒ f(3) = 2(3) − 1 = 6 − 1 = 5,

⇒ f(5) = 2(5) − 1 = 10 − 1 = 9.

The range of f is the set of all these images.

Hence, range of f = {−5, −1, 5, 9}.

From the given arrow diagram,

A = {−1, 0, 1, 2, 3} and B = {0, 2, 4, 7}

and the arrows join −1 to 2, 0 to 0, 1 to 2, 2 to 7 and 3 to 7.

Each of the five elements −1, 0, 1, 2 and 3 of A has exactly one arrow starting from it, so every element of A has one and only one image in B.

∴ The given arrow diagram represents a function.

The images are 2, 0, 2, 7 and 7, so the range is the set of these images.

Hence, the given arrow diagram represents a function whose range = {0, 2, 7}.

From the given arrow diagram,

the arrows join 1 to 4, 2 to 1, 3 to 4, 4 to 3 and 5 to 4.

Thus, the corresponding ordered pairs are (1, 4), (2, 1), (3, 4), (4, 3) and (5, 4).

∴ The relation = {(1, 4), (2, 1), (3, 4), (4, 3), (5, 4)}

The first components 1, 2, 3, 4 and 5 are all different, so each element of the domain has a unique image in the codomain.

Hence, the relation in roster form is {(1, 4), (2, 1), (3, 4), (4, 3), (5, 4)} and it is a function, because each element of the domain has a unique image in the codomain.

From the given arrow diagram,

A = {−1, 0, 1, 2, −2} and B = {0, 1, 4}

The arrows join −1 to 1, 0 to 0, 1 to 1, 2 to 4 and −2 to 4.

Thus, the corresponding ordered pairs are (−1, 1), (0, 0), (1, 1), (2, 4) and (−2, 4).

∴ The relation = {(−1, 1), (0, 0), (1, 1), (2, 4), (−2, 4)}

Each of the five elements of A has exactly one arrow starting from it, so every element of A has one and only one image in B.

∴ The given relation is a function.

Observing the pairs,

1 = (−1)², 0 = 0², 1 = 1², 4 = 2², 4 = (−2)²

so in each pair the second component is the square of the first component.

Hence, the relation is {(−1, 1), (0, 0), (1, 1), (2, 4), (−2, 4)}, it is a function and the rule is f(x) = x², x ∈ A.

From the given arrow diagram,

A = {p, q, r, s} and B = {1, 3, 5, 7}

and the arrows join p to 3, q to 7, r to 7 and s to 1.

Thus, the corresponding ordered pairs are (p, 3), (q, 7), (r, 7) and (s, 1).

∴ The relation = {(p, 3), (q, 7), (r, 7), (s, 1)}

The first components are p, q, r, s, so the domain is A.

The relation is from A to B, so the codomain is B.

The images are 3, 7, 7 and 1, so the range is the set of these images.

Each of the four elements p, q, r and s of A has exactly one arrow starting from it, so every element of A has one and only one image in B.

Hence, the relation is {(p, 3), (q, 7), (r, 7), (s, 1)}, domain = {p, q, r, s}, codomain = {1, 3, 5, 7}, range = {1, 3, 7} and the relation is a function.

Given,

$A = {-2, -1, 1, 2}$ and $f = \bigg\lbrace\left(x, \frac{1}{x}\right) : x \in A\bigg\rbrace$

Substituting each value of $x$,

when $x = -2$, $\frac{1}{x} = -\frac{1}{2}$,

when $x = -1$, $\frac{1}{x} = -1$,

when $x = 1$, $\frac{1}{x} = 1$,

when $x = 2$, $\frac{1}{x} = \frac{1}{2}$.

Thus, the corresponding ordered pairs are $\left(-2, -\frac{1}{2}\right)$, $(-1, -1)$, $(1, 1)$ and $\left(2, \frac{1}{2}\right)$.

∴ $f = \bigg\lbrace\left(-2, -\frac{1}{2}\right), (-1, -1), (1, 1), \left(2, \frac{1}{2}\right)\bigg\rbrace$

(i) The first components of the ordered pairs of $f$ are $-2$, $-1$, 1 and 2.

Hence, domain of $\bm{f = {-2, -1, 1, 2}}$.

(ii) The second components of the ordered pairs of $f$ are $-\frac{1}{2}$, $-1$, 1 and $\frac{1}{2}$.

Hence, range of $\bm{f = \bigg\lbrace-\frac{1}{2}, -1, 1, \frac{1}{2}\bigg\rbrace}$.

(iii) The first components $-2$, $-1$, 1 and 2 are all different, so every element of $A$ has one and only one image.

Hence, $\bm{f}$ is a function.

Given,

f(x) = x³ + 1, where X = {−1, 0, 3, 9, 7}

Substituting each value of x,

f(−1) = (−1)³ + 1 = −1 + 1 = 0,

f(0) = 0³ + 1 = 0 + 1 = 1,

f(3) = 3³ + 1 = 27 + 1 = 28,

f(9) = 9³ + 1 = 729 + 1 = 730,

f(7) = 7³ + 1 = 343 + 1 = 344.

Thus, the corresponding ordered pairs are (−1, 0), (0, 1), (3, 28), (9, 730) and (7, 344).

∴ f = {(−1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}

The range of f is the set of all these images.

Hence, f = {(−1, 0), (0, 1), (3, 28), (9, 730), (7, 344)} and range of f = {0, 1, 28, 730, 344}.

Given,

f(x) = ax + b

As f(−1) = −5,

⇒ a(−1) + b = −5

⇒ −a + b = −5 ...........(1)

As f(3) = 3,

⇒ a(3) + b = 3

⇒ 3a + b = 3 ...........(2)

Subtracting equation (1) from equation (2), we get :

⇒ 3a + b − (−a + b) = 3 − (−5)

⇒ 3a + b + a − b = 3 + 5

⇒ 4a = 8

⇒ a = 2

Substituting a = 2 in equation (1), we get :

⇒ −2 + b = −5

⇒ b = −5 + 2 = −3

Hence, a = 2 and b = −3.

Given,

$f : x \rightarrow 3x + 4$ i.e. $f(x) = 3x + 4$

As $(a, 8)$ belongs to $f$, $f(a) = 8$.

⇒ $3a + 4 = 8$

⇒ $3a = 8 - 4$

⇒ $3a = 4$

⇒ $a = \frac{4}{3}$

As $(2, b)$ belongs to $f$, $f(2) = b$.

⇒ $b = 3(2) + 4$

⇒ $b = 6 + 4$

⇒ $b = 10$

Hence, $\bm{a = \frac{4}{3}}$ and $\bm{b = 10}$.

Given,

f = {(x, 3x − 5) : x ∈ R} i.e. f(x) = 3x − 5

As (a, 4) belongs to f, f(a) = 4.

⇒ 3a − 5 = 4

⇒ 3a = 4 + 5

⇒ 3a = 9

⇒ a = 3

As (1, b) belongs to f, f(1) = b.

⇒ b = 3(1) − 5

⇒ b = 3 − 5

⇒ b = −2

Hence, a = 3 and b = −2.

Given,

$f(x) = x^2$

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

$$\begin{aligned}\therefore \frac{f(1.1) - f(1)}{1.1 - 1} = \frac{1.21 - 1}{1.1 - 1} \\[1em] = \frac{0.21}{0.1} \\[1em] = 2.1\end{aligned}$$

Hence, $\bm{\frac{f(1.1) - f(1)}{1.1 - 1} = 2.1}$.

Given,

$f(x) = \frac{3x^2 + 2x - 1}{x + 1}$, $x \ne -1$

As $-3 \in D_f$, $f(-3)$ exists.

Substituting $x = -3$, we get :

$f(-3) = \frac{3(-3)^2 + 2(-3) - 1}{-3 + 1}$

⇒ $f(-3) = \frac{27 - 6 - 1}{-2}$

⇒ $f(-3) = \frac{20}{-2}$

⇒ $f(-3) = -10$

∴ $f(-3) + 1 = -10 + 1 = -9$

Hence, $\bm{f(-3) + 1 = -9}$.

Given,

$$f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\[1em] 2x, & 3 \le x \le 5 \end{cases}$$

The value $x = 3$ lies in both the intervals $0 \le x \le 3$ and $3 \le x \le 5$, so both the rules apply to $x = 3$.

By the first rule,

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

By the second rule,

$f(3) = 2 \times 3 = 6$

Thus the element 3 of the domain of $f$ has two different images 9 and 6.

Hence, the given relation $\bm{f}$ is not a function, because the element 3 of its domain has two different images 9 and 6.

Given,

$f(x) = 3x - 2$, $D_f = \textbf{R}$

Substituting different values of $x$, we get :

⇒ $f(-3) = 3(-3) - 2 = -9 - 2 = -11$,

⇒ $f(0) = 3(0) - 2 = 0 - 2 = -2$,

⇒ $f(7) = 3(7) - 2 = 21 - 2 = 19$.

∴ $f(-3) + f(0) \times f(7) = -11 + (-2) \times 19$

⇒ $f(-3) + f(0) \times f(7) = -11 - 38$

⇒ $f(-3) + f(0) \times f(7) = -49$

Hence, $\bm{f(-3) + f(0) \times f(7) = -49}$.

Given,

$f(x) = x^3 - 1$ and $f(x) = 215$

⇒ $x^3 - 1 = 215$

⇒ $x^3 = 215 + 1$

⇒ $x^3 = 216$

⇒ $x^3 = 6^3$

⇒ $x = 6$

Hence, $\bm{x = 6}$.

Given,

$f(x) = 2x^2 + 1$ and $D_f = {-2, -1, 0, 1, 3}$

(i) Substituting each value of $x$ in $f(x) = 2x^2 + 1$, we get :

$f(-2) = 2(-2)^2 + 1 = 8 + 1 = 9$,

$f(-1) = 2(-1)^2 + 1 = 2 + 1 = 3$,

$f(0) = 2(0)^2 + 1 = 0 + 1 = 1$,

$f(1) = 2(1)^2 + 1 = 2 + 1 = 3$,

$f(3) = 2(3)^2 + 1 = 18 + 1 = 19$.

$R_f$ is the set of all these images.

Hence, range of $\bm{f = {1, 3, 9, 19}}$.

(ii) From part (i), $f(3) = 19$ and $f(-2) = 9$.

∴ $f(3) \times f(-2) = 19 \times 9 = 171$

Hence, $\bm{f(3) \times f(-2) = 171}$.

(iii) Given, $f(x) = 9$

⇒ $2x^2 + 1 = 9$

⇒ $2x^2 = 9 - 1$

⇒ $2x^2 = 8$

⇒ $x^2 = 4$

⇒ $x = \pm 2$

Out of 2 and $-2$, only $-2$ belongs to $D_f$.

Hence, $\bm{x = -2}$.

Given,

$f(x) = x^2 + 3$, $x \in \textbf{N}$ and $x \le 5$

The natural numbers not exceeding 5 are 1, 2, 3, 4 and 5.

∴ $D_f = {1, 2, 3, 4, 5}$

(i) Substituting each value of $x$ in $f(x) = x^2 + 3$, we get :

$f(1) = 1^2 + 3 = 1 + 3 = 4$,

$f(2) = 2^2 + 3 = 4 + 3 = 7$,

$f(3) = 3^2 + 3 = 9 + 3 = 12$,

$f(4) = 4^2 + 3 = 16 + 3 = 19$,

$f(5) = 5^2 + 3 = 25 + 3 = 28$.

$R_f$ is the set of all these images.

Hence, range of $\bm{f = {4, 7, 12, 19, 28}}$.

(ii) From part (i), $f(2) = 7$ and $f(5) = 28$.

∴ $f(2) \times f(5) = 7 \times 28 = 196$

Hence, $\bm{f(2) \times f(5) = 196}$.

(iii) $D_f = {1, 2, 3, 4, 5}$ and $-3$ does not belong to it, as $-3$ is not a natural number.

Hence, $\bm{f(-3)}$ does not exist.

(iv) Given, $f(x) = 7$

⇒ $x^2 + 3 = 7$

⇒ $x^2 = 7 - 3$

⇒ $x^2 = 4$

⇒ $x = \pm 2$

Out of 2 and $-2$, only 2 belongs to $D_f$.

Hence, $\bm{x = 2}$.

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

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{x^2 - 4}{x + 2}$ must be a real number

⇒ $x + 2 \ne 0$

⇒ $x \ne -2$

⇒ $D_f$ = set of all real numbers except $-2$ i.e. $D_f = \textbf{R} - {-2}$

Hence, domain of $\bm{f = \textbf{R} - {-2}}$.

(ii) $f(x) = \frac{3x}{28 - x}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{3x}{28 - x}$ must be a real number

⇒ $28 - x \ne 0$

⇒ $x \ne 28$

⇒ $D_f$ = set of all real numbers except 28 i.e. $D_f = \textbf{R} - {28}$

Hence, domain of $\bm{f = \textbf{R} - {28}}$.

(iii) $f(x) = \frac{x}{x^2 + 3x + 2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{x}{x^2 + 3x + 2}$ must be a real number

⇒ $x^2 + 3x + 2 \ne 0$

⇒ $x^2 + 2x + x + 2 \ne 0$

⇒ $x(x + 2) + 1(x + 2) \ne 0$

⇒ $(x + 2)(x + 1) \ne 0$

⇒ $x \ne -2, -1$

⇒ $D_f$ = set of all real numbers except $-2$ and $-1$ i.e. $D_f = \textbf{R} - {-1, -2}$

Hence, domain of $\bm{f = \textbf{R} - {-1, -2}}$.

(iv) $f(x) = \frac{x^3 - x + 3}{x^2 - 1}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{x^3 - x + 3}{x^2 - 1}$ must be a real number

⇒ $x^2 - 1 \ne 0$

⇒ $(x - 1)(x + 1) \ne 0$

⇒ $x \ne 1, -1$

⇒ $D_f$ = set of all real numbers except $-1$ and 1 i.e. $D_f = \textbf{R} - {-1, 1}$

Hence, domain of $\bm{f = \textbf{R} - {-1, 1}}$.

(i) $f(x) = x^2 + 2$, $x \in \textbf{R}$

For $R_f$, let $y = f(x) = x^2 + 2$, $x \in \textbf{R}$.

We know that for all $x \in \textbf{R}$, $x^2 \ge 0$

⇒ $x^2 + 2 \ge 0 + 2$

⇒ $y \ge 2$

⇒ $R_f = [2, \infty)$

Hence, range of $\bm{f = [2, \infty)}$.

(ii) $f(x) = 3 - 2x$, $x \in \textbf{R}$, $x \ge 1$

For $R_f$, let $y = f(x) = 3 - 2x$, $x \in \textbf{R}$, $x \ge 1$.

Now, $x \ge 1$

⇒ $2x \ge 2$

⇒ $-2x \le -2$

⇒ $3 - 2x \le 3 - 2$

⇒ $y \le 1$

⇒ $R_f = (-\infty, 1]$

Hence, range of $\bm{f = (-\infty, 1]}$.

Given,

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

For $D_f$, $f(x)$ must be a real number

⇒ $2x^2 + 1$ must be a real number, which is a real number for every $x \in \textbf{R}$

⇒ $D_f = \textbf{R}$

For $R_f$, let $y = f(x) = 2x^2 + 1$ ...........(1)

We know that for all $x \in \textbf{R}$, $x^2 \ge 0$

⇒ $2x^2 \ge 0$

⇒ $2x^2 + 1 \ge 0 + 1$

⇒ $y \ge 1$

⇒ $R_f = [1, \infty)$

Further, as $-2 \in D_f$, $f(-2)$ exists and $f(-2) = 2(-2)^2 + 1 = 8 + 1 = 9$.

As $51 \in R_f$, on putting $y = 51$ in equation (1), we get :

⇒ $2x^2 + 1 = 51$

⇒ $2x^2 = 51 - 1$

⇒ $2x^2 = 50$

⇒ $x^2 = 25$

⇒ $x = \pm 5$

Therefore, 5 and $-5$ are the numbers (in $D_f$) which are associated with the number 51 in $R_f$.

Hence, domain of $\bm{f = \textbf{R}}$, range of $\bm{f = [1, \infty)}$, $\bm{f(-2) = 9}$ and the numbers associated with 51 are 5 and $\bm{-5}$.

Given,

$f(x) = \sqrt{x + 2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\sqrt{x + 2}$ must be a real number

⇒ $x + 2 \ge 0$

⇒ $x \ge -2$

⇒ $D_f = [-2, \infty)$

For $R_f$, let $y = f(x) = \sqrt{x + 2}$.

As $x \ge -2$, $x + 2 \ge 0$

⇒ $\sqrt{x + 2} \ge 0$

⇒ $y \ge 0$

⇒ $R_f = [0, \infty)$

Hence, domain of $\bm{f = [-2, \infty)}$ and range of $\bm{f = [0, \infty)}$.

Given,

$f(x) = \sqrt{3 - 2x}$

For $D_f$, $f(x)$ must be a real number

⇒ $\sqrt{3 - 2x}$ must be a real number

⇒ $3 - 2x \ge 0$

⇒ $3 \ge 2x$

⇒ $x \le \frac{3}{2}$

⇒ $D_f = \left(-\infty, \frac{3}{2}\right]$

For $R_f$, let $y = f(x) = \sqrt{3 - 2x}$.

As $x \le \frac{3}{2}$, $2x \le 3$ i.e. $3 - 2x \ge 0$

⇒ $\sqrt{3 - 2x} \ge 0$

⇒ $y \ge 0$

⇒ $R_f = [0, \infty)$

Hence, domain of $\bm{f = \left(-\infty, \frac{3}{2}\right]}$ and range of $\bm{f = [0, \infty)}$.

Given,

$f(x) = \frac{1}{\sqrt{x - 5}}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{1}{\sqrt{x - 5}}$ must be a real number

⇒ $x - 5 \gt 0$

⇒ $x \gt 5$

⇒ $D_f = (5, \infty)$

For $R_f$, let $y = f(x) = \frac{1}{\sqrt{x - 5}}$.

As $x \gt 5$, $0 \lt x - 5$

⇒ $\sqrt{x - 5} \gt 0$

⇒ $\frac{1}{\sqrt{x - 5}} \gt 0$ $\left(\because \frac{1}{a} \gt 0 \text{ if and only if } a \gt 0\right)$

⇒ $y \gt 0$

⇒ $R_f = (0, \infty)$

Hence, domain of $\bm{f = (5, \infty)}$ and range of $\bm{f = (0, \infty)}$.

Given,

$f(x) = \frac{4 - x}{x - 4}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{4 - x}{x - 4}$ must be a real number

⇒ $x - 4 \ne 0$

⇒ $x \ne 4$

⇒ $D_f$ = set of all real numbers except 4 i.e. $D_f = \textbf{R} - {4}$

For $R_f$, let $y = \frac{4 - x}{x - 4}$.

For every $x \ne 4$,

⇒ $y = \frac{-(x - 4)}{x - 4}$

⇒ $y = -1$

So $f(x)$ takes only the value $-1$.

⇒ $R_f = {-1}$

Hence, domain of $\bm{f = \textbf{R} - {4}}$ and range of $\bm{f = {-1}}$.

Given,

$f(x) = \frac{x^2 - 9}{x - 3}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{x^2 - 9}{x - 3}$ must be a real number

⇒ $x - 3 \ne 0$

⇒ $x \ne 3$

⇒ $D_f$ = set of all real numbers except 3 i.e. $D_f = \textbf{R} - {3}$

For $R_f$, let $y = \frac{x^2 - 9}{x - 3}$.

For every $x \ne 3$,

⇒ $y = \frac{(x - 3)(x + 3)}{x - 3}$

⇒ $y = x + 3$

As $x$ takes every real value except 3, $x + 3$ takes every real value except $3 + 3 = 6$.

⇒ $R_f = \textbf{R} - {6}$

Hence, domain of $\bm{f = \textbf{R} - {3}}$ and range of $\bm{f = \textbf{R} - {6}}$.

Given,

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

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{x + 1}{2x + 1}$ must be a real number

⇒ $2x + 1 \ne 0$

⇒ $x \ne -\frac{1}{2}$

⇒ $D_f$ = set of all real numbers except $-\frac{1}{2}$ i.e. $D_f = \textbf{R} - \bigg\lbrace-\frac{1}{2}\bigg\rbrace$

For $R_f$, let $y = \frac{x + 1}{2x + 1}$

⇒ $y(2x + 1) = x + 1$

⇒ $2xy + y = x + 1$

⇒ $2xy - x = 1 - y$

⇒ $(2y - 1)x = 1 - y$

⇒ $x = \frac{1 - y}{2y - 1}$ but $x \in \textbf{R}$

⇒ $\frac{1 - y}{2y - 1}$ must be a real number

⇒ $2y - 1 \ne 0$

⇒ $y \ne \frac{1}{2}$

⇒ $R_f$ = set of all real numbers except $\frac{1}{2}$ i.e. $R_f = \textbf{R} - \bigg\lbrace\frac{1}{2}\bigg\rbrace$

Hence, domain of $\bm{f = \textbf{R} - \bigg\lbrace-\frac{1}{2}\bigg\rbrace}$ and range of $\bm{f = \textbf{R} - \bigg\lbrace\frac{1}{2}\bigg\rbrace}$.

Given,

$f(x) = \frac{4 + x}{4 - x}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{4 + x}{4 - x}$ must be a real number

⇒ $4 - x \ne 0$

⇒ $x \ne 4$

⇒ $D_f$ = set of all real numbers except 4 i.e. $D_f = \textbf{R} - {4}$

For $R_f$, let $y = \frac{4 + x}{4 - x}$

⇒ $y(4 - x) = 4 + x$

⇒ $4y - xy = 4 + x$

⇒ $4y - 4 = x + xy$

⇒ $(1 + y)x = 4y - 4$

⇒ $x = \frac{4y - 4}{y + 1}$ but $x \in \textbf{R}$

⇒ $\frac{4y - 4}{y + 1}$ must be a real number

⇒ $y + 1 \ne 0$

⇒ $y \ne -1$

⇒ $R_f$ = set of all real numbers except $-1$ i.e. $R_f = \textbf{R} - {-1}$

Hence, domain of $\bm{f = \textbf{R} - {4}}$ and range of $\bm{f = \textbf{R} - {-1}}$.

Given,

$f(x) = \frac{1}{4 - x^2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{1}{4 - x^2}$ must be a real number

⇒ $4 - x^2 \ne 0$

⇒ $(2 - x)(2 + x) \ne 0$

⇒ $x \ne \pm 2$

⇒ $D_f$ = set of all real numbers except $-2$ and 2 i.e. $D_f = \textbf{R} - {-2, 2}$

For $R_f$, let $y = \frac{1}{4 - x^2}$

⇒ $4 - x^2 = \frac{1}{y}$, $y \ne 0$

⇒ $x^2 = 4 - \frac{1}{y}$, $y \ne 0$

As $x^2 \ge 0$ for all $x \in D_f$

⇒ $4 - \frac{1}{y} \ge 0$, $y \ne 0$ but y^2 &ampgt 0

(Multiplying both the sides by $y^2$, a positive real number)

⇒ $y^2\left(4 - \frac{1}{y}\right) \ge 0$, $y \ne 0$

⇒ $y(4y - 1) \ge 0$, $y \ne 0$

⇒ $4y\left(y - \frac{1}{4}\right) \ge 0$, $y \ne 0$

⇒ either $y \le 0$ or $y \ge \frac{1}{4}$ but $y \ne 0$

⇒ $y \lt 0$ or $y \ge \frac{1}{4}$

⇒ $R_f = (-\infty, 0) \cup \left[\frac{1}{4}, \infty\right)$

Hence, domain of $\bm{f = \textbf{R} - {-2, 2}}$ and range of $\bm{f = (-\infty, 0) \cup \left[\frac{1}{4}, \infty\right)}$.

Given,

$f(x) = \sqrt{16 - x^2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\sqrt{16 - x^2}$ must be a real number

⇒ $16 - x^2 \ge 0$

⇒ $-(x^2 - 16) \ge 0$

⇒ $x^2 - 16 \le 0$

⇒ $(x + 4)(x - 4) \le 0$

⇒ $-4 \le x \le 4$

⇒ $D_f = [-4, 4]$

For $R_f$, let $y = \sqrt{16 - x^2}$ ...........(1)

As the square root of a real number is always non-negative, $y \ge 0$.

On squaring equation (1), we get :

⇒ $y^2 = 16 - x^2$

⇒ $x^2 = 16 - y^2$ but $x^2 \ge 0$ for all $x \in D_f$

⇒ $16 - y^2 \ge 0$

⇒ $y^2 - 16 \le 0$

⇒ $(y + 4)(y - 4) \le 0$

⇒ $-4 \le y \le 4$ but $y \ge 0$

⇒ $R_f = [0, 4]$

Hence, domain of $\bm{f = [-4, 4]}$ and range of $\bm{f = [0, 4]}$.

Given,

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

For $D_f$, $f(x)$ must be a real number

⇒ $\sqrt{x^2 - 9}$ must be a real number

⇒ $x^2 - 9 \ge 0$

⇒ $(x + 3)(x - 3) \ge 0$

⇒ either $x \le -3$ or $x \ge 3$

⇒ $D_f = (-\infty, -3] \cup [3, \infty)$

For $R_f$, let $y = \sqrt{x^2 - 9}$ ...........(1)

As the square root of a real number is always non-negative, $y \ge 0$.

On squaring equation (1), we get :

⇒ $y^2 = x^2 - 9$

⇒ $x^2 = y^2 + 9$ but $x^2 \ge 0$ for all $x \in D_f$

⇒ $y^2 + 9 \ge 0$

⇒ $y^2 \ge -9$, which is true for all $y \in \textbf{R}$. Also $y \ge 0$

⇒ $R_f = [0, \infty)$

Hence, domain of $\bm{f = (-\infty, -3] \cup [3, \infty)}$ and range of $\bm{f = [0, \infty)}$.

Given,

$f(x) = \frac{1}{\sqrt{4 - x^2}}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{1}{\sqrt{4 - x^2}}$ must be a real number

⇒ $4 - x^2 \gt 0$

⇒ $-(x^2 - 4) \gt 0$

⇒ $x^2 - 4 \lt 0$

⇒ $(x + 2)(x - 2) \lt 0$

⇒ $-2 \lt x \lt 2$

⇒ $D_f = (-2, 2)$

For $R_f$, let $y = \frac{1}{\sqrt{4 - x^2}}$, $y \ne 0$ ...........(1)

Also, as the square root of a real number is always non-negative, $y \gt 0$.

On squaring equation (1), we get :

⇒ $y^2 = \frac{1}{4 - x^2}$

⇒ $4 - x^2 = \frac{1}{y^2}$

⇒ $x^2 = 4 - \frac{1}{y^2}$

But $x^2 \ge 0$ for all $x \in D_f$

⇒ $4 - \frac{1}{y^2} \ge 0$ but y^2 &ampgt 0

(Multiplying both the sides by $y^2$, a positive real number)

⇒ $4y^2 - 1 \ge 0$

⇒ $y^2 - \frac{1}{4} \ge 0$

⇒ $\left(y + \frac{1}{2}\right)\left(y - \frac{1}{2}\right) \ge 0$

⇒ either $y \le -\frac{1}{2}$ or $y \ge \frac{1}{2}$ but $y \gt 0$

⇒ $y \ge \frac{1}{2}$

⇒ $R_f = \left[\frac{1}{2}, \infty\right)$

Hence, domain of $\bm{f = (-2, 2)}$ and range of $\bm{f = \left[\frac{1}{2}, \infty\right)}$.

(i) $f(x) = |x - 3|$

For $D_f$, $f(x)$ must be a real number

⇒ $|x - 3|$ must be a real number, which is a real number for every $x \in \textbf{R}$

⇒ $D_f = \textbf{R}$

For $R_f$, let $y = f(x) = |x - 3|$.

Since $|x| \ge 0$ for all $x \in \textbf{R}$

⇒ $|x - 3| \ge 0$ for all $x \in \textbf{R}$

⇒ $y \ge 0$

⇒ $R_f = [0, \infty)$

Hence, domain of $\bm{f = \textbf{R}}$ and range of $\bm{f = [0, \infty)}$.

(ii) $f(x) = 3 - |x - 2|$

For $D_f$, $f(x)$ must be a real number

⇒ $3 - |x - 2|$ must be a real number, which is a real number for every $x \in \textbf{R}$

⇒ $D_f = \textbf{R}$

For $R_f$, let $y = f(x) = 3 - |x - 2|$.

Since $|x| \ge 0$ for all $x \in \textbf{R}$

⇒ $|x - 2| \ge 0$ for all $x \in \textbf{R}$

⇒ $-|x - 2| \le 0$

⇒ $3 - |x - 2| \le 3$

⇒ $y \le 3$

⇒ $R_f = (-\infty, 3]$

Hence, domain of $\bm{f = \textbf{R}}$ and range of $\bm{f = (-\infty, 3]}$.

Given,

$f(x) = \frac{|x| - x}{2x}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{|x| - x}{2x}$ must be a real number

⇒ $2x \ne 0$

⇒ $x \ne 0$

⇒ $D_f$ = set of all real numbers except 0 i.e. $D_f = \textbf{R} - {0}$

For $R_f$, let $y = \frac{|x| - x}{2x}$. Two cases arise.

We know that,

$$|x| = \begin{cases} x, & \text{if } x \ge 0 \\[1em] -x, & \text{if } x \lt 0 \end{cases}$$

Case I. If $x \gt 0$, then $|x| = x$,

∴ $y = \frac{x - x}{2x} = \frac{0}{2x} = 0$.

Case II. If $x \lt 0$, then $|x| = -x$,

∴ $y = \frac{-x - x}{2x} = \frac{-2x}{2x} = -1$.

∴ On combining the two cases, $R_f = {0, -1}$.

Hence, domain of $\bm{f = \textbf{R} - {0}}$ and range of $\bm{f = {0, -1}}$.

Given,

$f(x) = \frac{|x - 4|}{x - 4}$

For $D_f$, $f(x)$ must be a real number

⇒ $\frac{|x - 4|}{x - 4}$ must be a real number

⇒ $x - 4 \ne 0$

⇒ $x \ne 4$

⇒ $D_f$ = set of all real numbers except 4 i.e. $D_f = \textbf{R} - {4}$

For $R_f$, let $y = \frac{|x - 4|}{x - 4}$. Two cases arise.

Case I. If $x - 4 \gt 0$, then $|x - 4| = x - 4$,

∴ $y = \frac{x - 4}{x - 4} = 1$.

Case II. If $x - 4 \lt 0$, then $|x - 4| = -(x - 4)$,

∴ $y = \frac{-(x - 4)}{x - 4} = -1$.

∴ On combining the two cases, $R_f = {1, -1}$.

Hence, domain of $\bm{f = \textbf{R} - {4}}$ and range of $\bm{f = {1, -1}}$.

Given,

$f(x) = 2x + 1$ and $g(x) = 4x - 7$

(i) As $f(x) = g(x)$

⇒ $2x + 1 = 4x - 7$

⇒ $1 + 7 = 4x - 2x$

⇒ $8 = 2x$

⇒ $x = 4$

Hence, $\bm{f(x) = g(x)}$ when $\bm{x = 4}$.

(ii) As $f(x) \lt g(x)$

⇒ $2x + 1 \lt 4x - 7$

⇒ $1 + 7 \lt 4x - 2x$

⇒ $8 \lt 2x$

⇒ $4 \lt x$

⇒ $x \gt 4$

Hence, $\bm{f(x) \lt g(x)}$ when $\bm{x \gt 4}$.

Given,

$f(x) = 3x^2 - 1$ and $g(x) = 3 + x$

As the functions $f$ and $g$ are equal, $f(x) = g(x)$

⇒ $3x^2 - 1 = 3 + x$

⇒ $3x^2 - x - 1 - 3 = 0$

⇒ $3x^2 - x - 4 = 0$

⇒ $3x^2 - 4x + 3x - 4 = 0$

⇒ $x(3x - 4) + 1(3x - 4) = 0$

⇒ $(3x - 4)(x + 1) = 0$

⇒ $3x - 4 = 0$ or $x + 1 = 0$

⇒ $x = \frac{4}{3}$ or $x = -1$

Hence, the domain for which the functions $\bm{f}$ and $\bm{g}$ are equal is $\bm{\bigg\lbrace -1, \frac{4}{3}\bigg\rbrace}$.

We know that,

$[\cdot]$ denotes the greatest integer function and for an integer $n$,

$[k] = n$ ⇔ $n \le k \lt n + 1$.

(i) $[2x - 3] = 5$

⇒ $5 \le 2x - 3 \lt 6$

⇒ $5 + 3 \le 2x \lt 6 + 3$

⇒ $8 \le 2x \lt 9$

⇒ $4 \le x \lt \frac{9}{2}$

Hence, $\bm{x \in \left[4, \frac{9}{2}\right)}$.

(ii) $[2x + 1] = -2$

⇒ $-2 \le 2x + 1 \lt -1$

⇒ $-2 - 1 \le 2x \lt -1 - 1$

⇒ $-3 \le 2x \lt -2$

⇒ $-\frac{3}{2} \le x \lt -1$

Hence, $\bm{x \in \left[-\frac{3}{2}, -1\right)}$.

(iii) $[2 - 3x] = 5$

⇒ $5 \le 2 - 3x \lt 6$

⇒ $5 - 2 \le -3x \lt 6 - 2$

⇒ $3 \le -3x \lt 4$

Dividing throughout by $-3$ and reversing the inequalities,

⇒ $-1 \ge x \gt -\frac{4}{3}$

⇒ $-\frac{4}{3} \lt x \le -1$

Hence, $\bm{x \in \left(-\frac{4}{3}, -1\right]}$.

Given,

$f(x) = \log\left(\frac{1 + x}{1 - x}\right)$

Taking L.H.S.,

$f(x) + f(y) = \log\left(\frac{1 + x}{1 - x}\right) + \log\left(\frac{1 + y}{1 - y}\right)$

We know that, $\log m + \log n = \log (mn)$.

⇒ $f(x) + f(y) = \log\left(\frac{1 + x}{1 - x} \times \frac{1 + y}{1 - y}\right)$

⇒ $f(x) + f(y) = \log\left[\frac{(1 + x)(1 + y)}{(1 - x)(1 - y)}\right]$ ...........(1)

Taking R.H.S.,

$f\left(\frac{x + y}{1 + xy}\right) = \log\left(\frac{1 + \frac{x + y}{1 + xy}}{1 - \frac{x + y}{1 + xy}}\right)$

⇒ $f\left(\frac{x + y}{1 + xy}\right) = \log\left(\frac{\frac{1 + xy + x + y}{1 + xy}}{\frac{1 + xy - x - y}{1 + xy}}\right)$

⇒ $f\left(\frac{x + y}{1 + xy}\right) = \log\left(\frac{1 + xy + x + y}{1 + xy - x - y}\right)$

⇒ $f\left(\frac{x + y}{1 + xy}\right) = \log\left[\frac{(1 + x) + y(1 + x)}{(1 - x) - y(1 - x)}\right]$

⇒ $f\left(\frac{x + y}{1 + xy}\right) = \log\left[\frac{(1 + x)(1 + y)}{(1 - x)(1 - y)}\right]$ ...........(2)

From equations (1) and (2), we get:

$f(x) + f(y) = f\left(\frac{x + y}{1 + xy}\right)$

Hence, proved that $\bm{f(x) + f(y) = f\left(\frac{x + y}{1 + xy}\right)}$.

Given,

$f(x) = \frac{9^x}{9^x + 3}$

Replacing $x$ by $1 - x$,

$f(1 - x) = \frac{9^{1 - x}}{9^{1 - x} + 3}$

⇒ $f(1 - x) = \frac{\frac{9}{9^x}}{\frac{9}{9^x} + 3}$

Multiplying the numerator and the denominator by $9^x$,

⇒ $f(1 - x) = \frac{9}{9 + 3 \times 9^x}$

⇒ $f(1 - x) = \frac{9}{3(3 + 9^x)}$

⇒ $f(1 - x) = \frac{3}{9^x + 3}$

∴ $f(x) + f(1 - x) = \frac{9^x}{9^x + 3} + \frac{3}{9^x + 3}$

⇒ $f(x) + f(1 - x) = \frac{9^x + 3}{9^x + 3}$

⇒ $f(x) + f(1 - x) = 1$

Hence, proved that $\bm{f(x) + f(1 - x) = 1}$.

We know that,

$\log k$ is a real number only when $k \gt 0$.

(i) $\log(4x - 3)$

For $\log(4x - 3)$ to be a real number,

⇒ $4x - 3 \gt 0$

⇒ $4x \gt 3$

⇒ $x \gt \frac{3}{4}$

Hence, domain $\bm{= \left(\frac{3}{4}, \infty\right)}$.

(ii) $\frac{1}{\log(9 - x^2)}$

For $\log(9 - x^2)$ to be a real number,

⇒ $9 - x^2 \gt 0$

⇒ $x^2 \lt 9$

⇒ $-3 \lt x \lt 3$

Also, the denominator must not be zero.

⇒ $\log(9 - x^2) \ne 0$

⇒ $9 - x^2 \ne 1$

⇒ $x^2 \ne 8$

⇒ $x \ne \pm 2\sqrt{2}$

Both $2\sqrt{2}$ and $-2\sqrt{2}$ lie between $-3$ and 3, so they must be excluded.

Hence, domain $\bm{= (-3, 3) - {-2\sqrt{2}, 2\sqrt{2}}}$.

Given,

$f(x) = |x| + |x - 2|$

We know that,

$$|x| = \begin{cases} x, & \text{if } x \ge 0 \\[1em] -x, & \text{if } x \lt 0 \end{cases}$$

and

$$|x - 2| = \begin{cases} x - 2, & \text{if } x \ge 2 \\[1em] -(x - 2), & \text{if } x \lt 2 \end{cases}$$

When $x \lt 0$, $f(x) = -x + (2 - x) = 2 - 2x$

When $0 \le x \lt 2$, $f(x) = x + (2 - x) = 2$

When $x \ge 2$, $f(x) = x + (x - 2) = 2x - 2$

∴ $f(x)$ can be written as

$$f(x) = \begin{cases} 2 - 2x, & x \lt 0 \\[1em] 2, & 0 \le x \lt 2 \\[1em] 2x - 2, & x \ge 2 \end{cases}$$

Plotting these three pieces, we get the following graph.

From the graph, the lowest value taken by $f(x)$ is 2 (taken along the whole segment from $x = 0$ to $x = 2$) and $f(x)$ increases without limit as $x$ moves away from this segment on either side.

Hence, range of $\bm{f = [2, \infty)}$.

Here $f(x)$ takes the value 3 for every real value of $x$, so the graph is a straight line parallel to the $x$-axis at a distance of 3 units above it.

Taking a few values of $x$,

when $x = -2$, $f(x) = 3$; when $x = -1$, $f(x) = 3$; when $x = 0$, $f(x) = 3$; when $x = 1$, $f(x) = 3$; when $x = 2$, $f(x) = 3$.

Plotting the points $(-2, 3)$, $(-1, 3)$, $(0, 3)$, $(1, 3)$, $(2, 3)$ and joining them, we get the following graph.

Hence, the graph of $\bm{f(x) = 3}$ is the horizontal straight line drawn above.

This is of the form $f(x) = mx + c$, so the graph is a straight line.

Taking a few values of $x$,

when $x = -1$, $f(x) = 3(-1) - 2 = -5$; when $x = 0$, $f(x) = -2$; when $x = 1$, $f(x) = 1$; when $x = 2$, $f(x) = 4$.

Plotting the points $(-1, -5)$, $(0, -2)$, $(1, 1)$, $(2, 4)$ and joining them, we get a straight line.

Hence, the graph of $\bm{f(x) = 3x - 2}$ is the straight line drawn above.

Taking a few values of $x$,

when $x = -2$, $f(x) = (-2)^2 + 1 = 5$; when $x = -1$, $f(x) = 2$; when $x = 0$, $f(x) = 1$; when $x = 1$, $f(x) = 2$; when $x = 2$, $f(x) = 5$.

Plotting the points $(-2, 5)$, $(-1, 2)$, $(0, 1)$, $(1, 2)$, $(2, 5)$ and joining them by a free-hand smooth curve, we get a parabola opening upwards with its lowest point at $(0, 1)$.

Hence, the graph of $\bm{f(x) = x^2 + 1}$ is the parabola drawn above.

Taking a few values of $x$,

when $x = -2$, $f(x) = 2^{-2} = \frac{1}{4}$; when $x = -1$, $f(x) = \frac{1}{2}$; when $x = 0$, $f(x) = 1$; when $x = 1$, $f(x) = 2$; when $x = 2$, $f(x) = 4$.

Plotting the points $\left(-2, \frac{1}{4}\right)$, $\left(-1, \frac{1}{2}\right)$, $(0, 1)$, $(1, 2)$, $(2, 4)$ and joining them by a free-hand smooth curve, we get an increasing curve which lies wholly above the $x$-axis and passes through $(0, 1)$.

Hence, the graph of $\bm{f(x) = 2^x}$ is the exponential curve drawn above.

Taking a few values of $x$,

when $x = -2$, $f(x) = \frac{-2}{1 + 4} = -\frac{2}{5}$; when $x = -1$, $f(x) = -\frac{1}{2}$; when $x = 0$, $f(x) = 0$; when $x = 1$, $f(x) = \frac{1}{2}$; when $x = 2$, $f(x) = \frac{2}{5}$.

As $f(-x) = -f(x)$, the curve is symmetrical about the origin. It passes through the origin, rises to the highest point $\left(1, \frac{1}{2}\right)$ and then falls towards the $x$-axis.

Hence, the graph of $\bm{f(x) = \frac{x}{1 + x^2}}$ is the curve drawn above.

Given,

$f(x) = x^2 + 7$ and $g(x) = 3x + 5$

⇒ $f(3) = 3^2 + 7 = 9 + 7 = 16$

⇒ $g(-5) = 3(-5) + 5 = -15 + 5 = -10$

∴ $f(3) + g(-5) = 16 + (-10) = 6$

Hence, $\bm{f(3) + g(-5) = 6}$.

Given,

$f(x) = x^2 + 7$ and $g(x) = 3x + 5$

⇒ $f(-2) = (-2)^2 + 7 = 4 + 7 = 11$

⇒ $g(-1) = 3(-1) + 5 = -3 + 5 = 2$

∴ $f(-2) + g(-1) = 11 + 2 = 13$

Hence, $\bm{f(-2) + g(-1) = 13}$.

Given,

$f(x) = x^2 + 7$ and $g(x) = 3x + 5$

⇒ $f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + 7 = \frac{1}{4} + 7 = \frac{1 + 28}{4} = \frac{29}{4}$

⇒ $g(14) = 3(14) + 5 = 42 + 5 = 47$

∴ $f\left(\frac{1}{2}\right) \times g(14) = \frac{29}{4} \times 47 = \frac{1363}{4}$

Hence, $\bm{f\left(\frac{1}{2}\right) \times g(14) = \frac{1363}{4}}$.

Given,

$f(x) = x^2 + 7$

⇒ $f(t) = t^2 + 7$

⇒ $f(-2) = (-2)^2 + 7 = 4 + 7 = 11$

∴ $f(t) - f(-2) = t^2 + 7 - 11 = t^2 - 4$

Hence, $\bm{f(t) - f(-2) = t^2 - 4}$.

Given,

$f(x) = x^2 + 7$

⇒ $f(t) = t^2 + 7$

⇒ $f(5) = 5^2 + 7 = 25 + 7 = 32$

∴ $\frac{f(t) - f(5)}{t - 5} = \frac{t^2 + 7 - 32}{t - 5}$

⇒ $\frac{f(t) - f(5)}{t - 5} = \frac{t^2 - 25}{t - 5}$

⇒ $\frac{f(t) - f(5)}{t - 5} = \frac{(t - 5)(t + 5)}{t - 5}$

⇒ $\frac{f(t) - f(5)}{t - 5} = t + 5$ (as $t \ne 5$)

Hence, $\bm{\frac{f(t) - f(5)}{t - 5} = t + 5}$.

Given,

$f(x) = e^x$ and $g(x) = \log x$

We know that,

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

⇒ $(f - g)(1) = f(1) - g(1)$

⇒ $(f - g)(1) = e^1 - \log 1$

We know that, $\log 1 = 0$.

⇒ $(f - g)(1) = e - 0 = e$

Hence, $\bm{(f - g)(1) = e}$.

Given,

$f(x) = e^x$ and $g(x) = \log x$

We know that,

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

⇒ $(fg)(1) = f(1) \times g(1)$

⇒ $(fg)(1) = e^1 \times \log 1$

We know that, $\log 1 = 0$.

⇒ $(fg)(1) = e \times 0 = 0$

Hence, $\bm{(fg)(1) = 0}$.

Given,

$f(x) = e^x$ and $g(x) = \log x$

We know that,

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

⇒ $\left(\frac{f}{g}\right)(3) = \frac{f(3)}{g(3)}$

⇒ $\left(\frac{f}{g}\right)(3) = \frac{e^3}{\log 3}$

Hence, $\bm{\left(\frac{f}{g}\right)(3) = \frac{e^3}{\log 3}}$.

Given,

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

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(f + g)(x) = (2x + 1) + (x^2 + 1)$

⇒ $(f + g)(x) = x^2 + 2x + 2$

Hence, $\bm{(f + g)(x) = x^2 + 2x + 2}$, $\bm{x \in \textbf{R}}$.

Given,

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

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(f - g)(x) = (2x + 1) - (x^2 + 1)$

⇒ $(f - g)(x) = 2x + 1 - x^2 - 1$

⇒ $(f - g)(x) = 2x - x^2$

Hence, $\bm{(f - g)(x) = 2x - x^2}$, $\bm{x \in \textbf{R}}$.

Given,

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

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(fg)(x) = (2x + 1)(x^2 + 1)$

⇒ $(fg)(x) = 2x^3 + 2x + x^2 + 1$

⇒ $(fg)(x) = 2x^3 + x^2 + 2x + 1$

Hence, $\bm{(fg)(x) = 2x^3 + x^2 + 2x + 1}$, $\bm{x \in \textbf{R}}$.

Given,

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

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

Here $x^2 + 1 \ge 1$ for every real number $x$, so $g(x)$ is never zero.

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

Hence, $\bm{\left(\frac{f}{g}\right)(x) = \frac{2x + 1}{x^2 + 1}}$, $\bm{x \in \textbf{R}}$.

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(f + g)(x) = (x^3 + 1) + (x + 1)$

⇒ $(f + g)(x) = x^3 + x + 2$

Hence, $\bm{(f + g)(x) = x^3 + x + 2}$, $\bm{x \in \textbf{R}}$.

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(g - f)(x) = (x + 1) - (x^3 + 1)$

⇒ $(g - f)(x) = x + 1 - x^3 - 1$

⇒ $(g - f)(x) = x - x^3$

Hence, $\bm{(g - f)(x) = x - x^3}$, $\bm{x \in \textbf{R}}$.

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(fg)(x) = (x^3 + 1)(x + 1)$

⇒ $(fg)(x) = x^4 + x^3 + x + 1$

Hence, $\bm{(fg)(x) = x^4 + x^3 + x + 1}$, $\bm{x \in \textbf{R}}$.

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

Here $g(x) = x + 1 = 0$ when $x = -1$, so $x = -1$ must be excluded.

⇒ $\left(\frac{f}{g}\right)(x) = \frac{x^3 + 1}{x + 1}$

Hence, $\bm{\left(\frac{f}{g}\right)(x) = \frac{x^3 + 1}{x + 1}}$, $\bm{x \in \textbf{R}}$, $\bm{x \ne -1}$.

Given,

$f(x) = x^3 + 1$ and $g(x) = x + 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(2g^2 - 3f)(x) = 2[g(x)]^2 - 3f(x)$

⇒ $(2g^2 - 3f)(x) = 2(x + 1)^2 - 3(x^3 + 1)$

⇒ $(2g^2 - 3f)(x) = 2(x^2 + 2x + 1) - 3x^3 - 3$

⇒ $(2g^2 - 3f)(x) = 2x^2 + 4x + 2 - 3x^3 - 3$

⇒ $(2g^2 - 3f)(x) = 2x^2 + 4x - 3x^3 - 1$

Hence, $\bm{(2g^2 - 3f)(x) = 2x^2 + 4x - 3x^3 - 1}$, $\bm{x \in \textbf{R}}$.

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

For $f(x)$ to be a real number, $x - 2 \ge 0$ i.e. $x \ge 2$.

∴ $D_f = [2, \infty)$

For $g(x)$ to be a real number, $x^2 - 1 \ge 0$ i.e. $x \le -1$ or $x \ge 1$.

∴ $D_g = (-\infty, -1] \cup [1, \infty)$

The set common to both the domains is $[2, \infty)$.

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

⇒ $(f + g)(x) = \sqrt{x - 2} + \sqrt{x^2 - 1}$

Hence, $\bm{(f + g)(x) = \sqrt{x - 2} + \sqrt{x^2 - 1}}$, $\bm{x \in [2, \infty)}$.

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

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

⇒ $(g - f)(x) = \sqrt{x^2 - 1} - \sqrt{x - 2}$

Hence, $\bm{(g - f)(x) = \sqrt{x^2 - 1} - \sqrt{x - 2}}$, $\bm{x \in [2, \infty)}$.

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

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

⇒ $(fg)(x) = \sqrt{x - 2} \times \sqrt{x^2 - 1}$

⇒ $(fg)(x) = \sqrt{(x - 2)(x^2 - 1)}$

Hence, $\bm{(fg)(x) = \sqrt{(x - 2)(x^2 - 1)}}$, $\bm{x \in [2, \infty)}$.

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$(3f - 2g)(x) = 3f(x) - 2g(x)$

⇒ $(3f - 2g)(x) = 3\sqrt{x - 2} - 2\sqrt{x^2 - 1}$

Hence, $\bm{(3f - 2g)(x) = 3\sqrt{x - 2} - 2\sqrt{x^2 - 1}}$, $\bm{x \in [2, \infty)}$.

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

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

For $x \in [2, \infty)$, $x^2 - 1 \ge 3 \gt 0$, so $g(x)$ is never zero.

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

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

Hence, $\bm{\left(\frac{f}{g}\right)(x) = \sqrt{\frac{x - 2}{x^2 - 1}}}$, $\bm{x \in [2, \infty)}$.

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$(2f^2 + \sqrt{3}g)(x) = 2[f(x)]^2 + \sqrt{3},g(x)$

⇒ $(2f^2 + \sqrt{3}g)(x) = 2\left(\sqrt{x - 2}\right)^2 + \sqrt{3}\sqrt{x^2 - 1}$

⇒ $(2f^2 + \sqrt{3}g)(x) = 2(x - 2) + \sqrt{3(x^2 - 1)}$

Hence, $\bm{(2f^2 + \sqrt{3}g)(x) = 2(x - 2) + \sqrt{3(x^2 - 1)}}$, $\bm{x \in [2, \infty)}$.

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

$\left(\frac{1}{f}\right)(x) = \frac{1}{f(x)}$, where $f(x) \ne 0$.

Here $f(x) = \sqrt{x - 2} = 0$ when $x = 2$, so $x = 2$ must be excluded.

⇒ $\left(\frac{1}{f}\right)(x) = \frac{1}{\sqrt{x - 2}}$

Hence, $\bm{\left(\frac{1}{f}\right)(x) = \frac{1}{\sqrt{x - 2}}}$, $\bm{x \in (2, \infty)}$.

Given,

$f(x) = \sqrt{x - 2}$ and $g(x) = \sqrt{x^2 - 1}$

As shown in part (i), $D_f = [2, \infty)$, $D_g = (-\infty, -1] \cup [1, \infty)$ and the set common to both the domains is $[2, \infty)$.

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

Here $f(x) = \sqrt{x - 2} = 0$ when $x = 2$, so $x = 2$ must be excluded.

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

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

Hence, $\bm{\left(\frac{g}{f}\right)(x) = \sqrt{\frac{x^2 - 1}{x - 2}}}$, $\bm{x \in (2, \infty)}$.

Given,

$f(x) = x^3 - \frac{1}{x^3}$, where $x \ne 0$.

Replacing $x$ by $\frac{1}{x}$,

$f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^3 - \frac{1}{\left(\frac{1}{x}\right)^3}$

⇒ $f\left(\frac{1}{x}\right) = \frac{1}{x^3} - x^3$

∴ $f(x) + f\left(\frac{1}{x}\right) = \left(x^3 - \frac{1}{x^3}\right) + \left(\frac{1}{x^3} - x^3\right)$

⇒ $f(x) + f\left(\frac{1}{x}\right) = x^3 - \frac{1}{x^3} + \frac{1}{x^3} - x^3$

⇒ $f(x) + f\left(\frac{1}{x}\right) = 0$

Hence, proved that $\bm{f(x) + f\left(\frac{1}{x}\right) = 0}$.

Given,

$f(x) = x + \frac{1}{x}$, where $x \ne 0$.

Replacing $x$ by $x^3$,

$f(x^3) = x^3 + \frac{1}{x^3}$ ...........(1)

Replacing $x$ by $\frac{1}{x}$,

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

⇒ $f\left(\frac{1}{x}\right) = x + \frac{1}{x}$ ...........(2)

Taking L.H.S.,

$(f(x))^3 = \left(x + \frac{1}{x}\right)^3$

We know that, $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.

⇒ $(f(x))^3 = x^3 + \frac{1}{x^3} + 3 \times x \times \frac{1}{x}\left(x + \frac{1}{x}\right)$

⇒ $(f(x))^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right)$

Using equations (1) and (2),

⇒ $(f(x))^3 = f(x^3) + 3f\left(\frac{1}{x}\right)$

Hence, proved that $\bm{(f(x))^3 = f(x^3) + 3f\left(\frac{1}{x}\right)}$.

Given,

$y = f(x) = \frac{6x - 5}{5x - 6}$, where $x \ne \frac{6}{5}$.

Replacing $x$ by $y$ in the given function,

$f(y) = \frac{6y - 5}{5y - 6}$

Substituting $y = \frac{6x - 5}{5x - 6}$,

⇒ $f(y) = \frac{6\left(\frac{6x - 5}{5x - 6}\right) - 5}{5\left(\frac{6x - 5}{5x - 6}\right) - 6}$

⇒ $f(y) = \frac{\frac{6(6x - 5) - 5(5x - 6)}{5x - 6}}{\frac{5(6x - 5) - 6(5x - 6)}{5x - 6}}$

⇒ $f(y) = \frac{6(6x - 5) - 5(5x - 6)}{5(6x - 5) - 6(5x - 6)}$

⇒ $f(y) = \frac{36x - 30 - 25x + 30}{30x - 25 - 30x + 36}$

⇒ $f(y) = \frac{11x}{11}$

⇒ $f(y) = x$

Hence, proved that $\bm{f(y) = x}$.

Given,

A = {−1, 2, 5, 8} and B = {0, 1, 3, 6, 7}

R is the relation "is one less than" from A to B i.e. R = {(x, y) : x ∈ A, y ∈ B and y = x + 1}.

Taking each element of A in turn,

when x = −1, y = 0 ∈ B,

when x = 2, y = 3 ∈ B,

when x = 5, y = 6 ∈ B,

when x = 8, y = 9 ∉ B.

Thus, the corresponding ordered pairs are (−1, 0), (2, 3) and (5, 6).

∴ R = {(−1, 0), (2, 3), (5, 6)}

Hence, Option 1 is the correct option.

Given,

R = {(x, y) : x, y ∈ W, 2x + y = 8}

2x + y = 8 ⇒ y = 8 − 2x

Taking those whole numbers x for which y = 8 − 2x is also a whole number,

when x = 0, y = 8 ∈ W,

when x = 1, y = 6 ∈ W,

when x = 2, y = 4 ∈ W,

when x = 3, y = 2 ∈ W,

when x = 4, y = 0 ∈ W,

when x = 5, y = −2 ∉ W.

For all the values of x greater than 4, y = 8 − 2x is negative and so does not belong to W.

Thus, the corresponding ordered pairs are (0, 8), (1, 6), (2, 4), (3, 2) and (4, 0).

∴ R = {(0, 8), (1, 6), (2, 4), (3, 2), (4, 0)}

∴ Domain of R = {0, 1, 2, 3, 4}

Hence, Option 3 is the correct option.

Given,

R = {(x, y) : x, y ∈ N, x + 2y = 21}

x + 2y = 21 ⇒ x = 21 − 2y

Taking those natural numbers y for which x = 21 − 2y is also a natural number,

when y = 1, x = 19 ∈ N,

when y = 2, x = 17 ∈ N,

......

when y = 10, x = 1 ∈ N,

when y = 11, x = −1 ∉ N.

For all the values of y greater than 10, x = 21 − 2y is negative and so does not belong to N.

So y takes the values 1, 2, 3, ......, 10 and the range of R is the set of these values of y.

∴ Range of R = {1, 2, 3, ......, 9, 10}

Hence, Option 2 is the correct option.

Given,

A = {1, 2, 3, 4, 5, 6} and R = {(x, y) : x, y ∈ A, x divides y}

The range of R is the set of all those elements y of A which are divisible by at least one element x of A.

Every element y of A is divisible by 1 ∈ A, so (1, y) ∈ R for every y ∈ A.

∴ Every element of A occurs as a second component.

∴ Range of R = {1, 2, 3, 4, 5, 6}

Hence, Option 4 is the correct option.

Given,

A = {3, 5} and B = {7, 11} and R = {(a, b) : a ∈ A, b ∈ B, a − b is odd}

All the elements of A and B are odd numbers, and the difference of two odd numbers is always even.

Checking each pair,

3 − 7 = −4, 3 − 11 = −8, 5 − 7 = −2, 5 − 11 = −6

and each of these is even, so no ordered pair satisfies the given condition.

∴ R = φ

Hence, Option 2 is the correct option.

Given,

R = {(x, y) : x, y ∈ Z, y = x − 3}

An ordered pair (x, y) belongs to R only when y = x − 3.

For (1, 4), x − 3 = 1 − 3 = −2 ≠ 4

For (0, 3), x − 3 = 0 − 3 = −3 ≠ 3

For (5, 2), x − 3 = 5 − 3 = 2, which is the second component.

For (−4, 1), x − 3 = −4 − 3 = −7 ≠ 1

∴ Only the ordered pair (5, 2) belongs to R.

Hence, Option 3 is the correct option.

Given,

R = {(x, y) : x, y ∈ W, x² + y² = 169}

As x and y are whole numbers, neither x² nor y² can exceed 169, so x can take the values 0, 1, 2, ....., 13 only.

Taking those values of x for which y² = 169 − x² is the square of a whole number,

when x = 0,

⇒ y² = 169

⇒ y = 13 ∈ W

when x = 5,

⇒ y² = 169 − 25 = 144

⇒ y = 12 ∈ W

when x = 12,

⇒ y² = 169 − 144 = 25

⇒ y = 5 ∈ W

when x = 13,

⇒ y² = 169 − 169 = 0

⇒ y = 0 ∈ W

For all the other values of x, 169 − x² is not the square of a whole number.

Thus, the corresponding ordered pairs are (0, 13), (5, 12), (12, 5) and (13, 0).

∴ R = {(0, 13), (5, 12), (12, 5), (13, 0)}

∴ Domain of R = {0, 5, 12, 13}

Hence, Option 1 is the correct option.

Given,

A = {a, b} and B = {x, y, z}

∴ n(A) = 2 and n(B) = 3

We know that,

Number of relations from B to A = 2^{n(B) × n(A)}

∴ Number of relations from B to A = 2^{3 × 2} = 2⁶ = 64

Hence, Option 4 is the correct option.

Given,

n(A) = m and n(B) = n

∴ n(A × B) = n(A) × n(B) = mn

A relation from A to B is a subset of A × B, so the total number of relations from A to B is the number of subsets of A × B i.e. 2^mn.

Out of these, exactly one relation, namely the empty relation φ, is not a non-empty relation.

∴ Number of non-empty relations from A to B = 2^mn − 1

Hence, Option 3 is the correct option.

Given,

A is a finite set containing n elements i.e. n(A) = n.

A relation on A is a relation from A to A, so it is a subset of A × A.

∴ n(A × A) = n(A) × n(A) = n × n = n²

∴ Number of relations on A = number of subsets of A × A = 2ⁿ²

Hence, Option 2 is the correct option.

Given,

A = {2, 3, 4, 5, 6} and R = {(x, y) : y = x + 2, x, y ∈ A}

Taking those elements x of A for which y = x + 2 also belongs to A,

when x = 2, y = 4 ∈ A,

when x = 3, y = 5 ∈ A,

when x = 4, y = 6 ∈ A,

when x = 5, y = 7 ∉ A,

when x = 6, y = 8 ∉ A.

Thus, the corresponding ordered pairs are (2, 4), (3, 5) and (4, 6).

∴ R = {(2, 4), (3, 5), (4, 6)}

Hence, Option 1 is the correct option.

We know that,

A relation is a function if and only if no two different ordered pairs of the relation have the same first component.

In option 1, the pairs (3, 9) and (3, 11) have the same first component 3, so it is not a function.

In option 2, the pairs (2, 4) and (2, 6) have the same first component 2, so it is not a function.

In option 4, the pairs (3, 1) and (3, 5) have the same first component 3, so it is not a function.

In option 3, the first components 2, 4, 6, 8 and 10 are all different, so each element of the domain has a unique image.

Hence, Option 3 is the correct option.

We know that,

An arrow diagram represents a function from X to Y if and only if exactly one arrow starts from each element of X.

In the diagram of option 2, two arrows start from the element a of X, so the element a has two different images.

In the diagram of option 3, no arrow starts from the element c of X, so the element c has no image.

In the diagram of option 4, two arrows start from the element a of X and no arrow starts from the element b of X.

In the diagram of option 1, exactly one arrow starts from each of the elements a, b, c and d of X, so every element of X has one and only one image in Y.

Hence, Option 1 is the correct option.

Let n(A) = p and n(B) = q.

Each element of A can be associated with any one of the q elements of B, so each element of A can be given an image in q ways.

As A has p elements and the choice for one element is independent of the choice for another,

Total number of functions from A to B = q × q × q ...... p times = q^p

∴ Number of functions from A to B = {n(B)}^n(A)

Hence, Option 4 is the correct option.

Given,

n(A) = 3

We know that,

Number of functions from a set A to a set B = {n(B)}^n(A)

Here both the sets are A, so

Number of functions from A to A = {n(A)}^n(A) = 3³ = 27

Hence, Option 3 is the correct option.

Given,

$f(x) = \sqrt{a^2 - x^2}$, where $a \gt 0$.

For $f(x)$ to be a real number,

⇒ $a^2 - x^2 \ge 0$

⇒ $x^2 \le a^2$

⇒ $-a \le x \le a$

∴ $D_f = [-a, a]$

Hence, Option 2 is the correct option.

Given,

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

For $f(x)$ to be a real number,

⇒ $x^2 - 9 \ge 0$

⇒ $x^2 \ge 9$

⇒ $x \le -3$ or $x \ge 3$

∴ $D_f = (-\infty, -3] \cup [3, \infty)$

Hence, Option 3 is the correct option.

Given,

$f(x) = \frac{x^2 + 2x + 1}{x^2 - x - 6}$

$f(x)$ is not defined when the denominator is zero.

⇒ $x^2 - x - 6 = 0$

⇒ $x^2 - 3x + 2x - 6 = 0$

⇒ $x(x - 3) + 2(x - 3) = 0$

⇒ $(x - 3)(x + 2) = 0$

⇒ $x = 3$ or $x = -2$

∴ $D_f = \textbf{R} - {3, -2}$

Hence, Option 3 is the correct option.

Given,

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

For $f(x)$ to be a real number, $\sqrt{|x| - x}$ must be a real number and it must not be zero.

⇒ $|x| - x \gt 0$

⇒ $|x| \gt x$

When $x \ge 0$, $|x| = x$, so $|x| - x = 0$ which is not greater than 0.

When $x \lt 0$, $|x| = -x$, so $|x| - x = -x - x = -2x \gt 0$.

∴ $D_f = (-\infty, 0)$

Hence, Option 1 is the correct option.

Given,

$f(x) = \frac{1}{4x^2 - 1}$

$f(x)$ is not defined when the denominator is zero.

⇒ $4x^2 - 1 = 0$

⇒ $(2x - 1)(2x + 1) = 0$

⇒ $x = \frac{1}{2}$ or $x = -\frac{1}{2}$

∴ $D_f = \textbf{R} - \bigg\lbrace -\frac{1}{2}, \frac{1}{2} \bigg\rbrace$

For $R_f$, let $y = f(x) = \frac{1}{4x^2 - 1}$

⇒ $4x^2 - 1 = \frac{1}{y}$, $y \ne 0$

⇒ $x^2 = \frac{1}{4}\left(1 + \frac{1}{y}\right)$

For $x$ to be a real number,

⇒ $\frac{1}{4}\left(\frac{y + 1}{y}\right) \ge 0$

⇒ $\frac{y + 1}{y} \ge 0$

⇒ $y \le -1$ or $y \gt 0$

∴ $R_f = (-\infty, -1] \cup (0, \infty)$

Hence, Option 2 is the correct option.

Given,

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

For $f(x)$ to be a real number,

⇒ $x - 1 \ge 0$

⇒ $x \ge 1$

∴ $D_f = [1, \infty)$

For $R_f$, let $y = f(x) = \sqrt{x - 1}$.

As $x \ge 1$, $x - 1 \ge 0$

⇒ $\sqrt{x - 1} \ge 0$

⇒ $y \ge 0$

∴ $R_f = [0, \infty)$

Hence, Option 3 is the correct option.

Given,

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

$f(x)$ is not defined when the denominator is zero.

⇒ $2 - x = 0$

⇒ $x = 2$

∴ $D_f = \textbf{R} - {2}$

For every $x \ne 2$,

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

So $f(x)$ takes only the value $-1$.

∴ $R_f = {-1}$

Hence, Option 1 is the correct option.

Given,

$f(x) = \frac{x}{|x|}$

$f(x)$ is not defined when the denominator is zero i.e. when $|x| = 0$ i.e. when $x = 0$.

∴ $D_f = \textbf{R} - {0}$

When $x \gt 0$, $|x| = x$

⇒ $f(x) = \frac{x}{x} = 1$

When $x \lt 0$, $|x| = -x$

⇒ $f(x) = \frac{x}{-x} = -1$

So $f(x)$ takes only the values 1 and $-1$.

∴ $R_f = {-1, 1}$

Hence, Option 3 is the correct option.

Given,

f(x) = 2 − | x − 5 |

| x − 5 | is defined for every real value of x, so 2 − | x − 5 | is a real number for every real value of x.

∴ Domain of f = R

Let y = f(x), then

y = 2 − | x − 5 |

We know that | x − 5 | ≥ 0.

⇒ −| x − 5 | ≤ 0

⇒ 2 − | x − 5 | ≤ 2

⇒ y ≤ 2

∴ Range of f = (−∞, 2]

Hence, Option 2 is the correct option.

Given,

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

For $\sqrt{a - x}$ to be a real number,

⇒ $a - x \ge 0$

⇒ $x \le a$ ...........(1)

For $\frac{1}{\sqrt{x^2 - a^2}}$ to be a real number, $\sqrt{x^2 - a^2}$ must be a real number and it must not be zero.

⇒ $x^2 - a^2 \gt 0$

⇒ $x \lt -a$ or $x \gt a$ ...........(2)

The values of $x$ satisfying both the equations (1) and (2) are given by $x \lt -a$.

∴ $D_f = (-\infty, -a)$

Hence, Option 3 is the correct option.

Given,

$f(x) = \log_e (5 - 6x)$

We know that, $\log_e k$ is a real number only when k &ampgt 0.

⇒ $5 - 6x \gt 0$

⇒ $5 \gt 6x$

⇒ $x \lt \frac{5}{6}$

∴ $D_f = \left(-\infty, \frac{5}{6}\right)$

Hence, Option 1 is the correct option.

Given,

$f(x) = \frac{1}{4 - x^2} + \log_{10}(x^2 - x)$

For $\frac{1}{4 - x^2}$ to be defined, the denominator must not be zero.

⇒ $4 - x^2 \ne 0$

⇒ $x \ne \pm 2$ ...........(1)

For $\log_{10}(x^2 - x)$ to be a real number,

⇒ $x^2 - x \gt 0$

⇒ $x(x - 1) \gt 0$

⇒ $x \lt 0$ or $x \gt 1$ ...........(2)

From equation (2), $x \in (-\infty, 0) \cup (1, \infty)$ and from equation (1), the values $-2$ and 2 must be excluded from it.

∴ $D_f = (-\infty, 0) \cup (1, \infty) - {-2, 2}$

Hence, Option 2 is the correct option.

Given,

[x]² − 3[x] + 2 = 0

⇒ [x]² − 2[x] − [x] + 2 = 0

⇒ [x]([x] − 2) − 1([x] − 2) = 0

⇒ ([x] − 2)([x] − 1) = 0

⇒ [x] = 2 or [x] = 1

We know that, [x] = n ⇔ n ≤ x < n + 1.

When [x] = 1, 1 ≤ x < 2 i.e. x ∈ [1, 2)

When [x] = 2, 2 ≤ x < 3 i.e. x ∈ [2, 3)

∴ x ∈ [1, 2) ∪ [2, 3) = [1, 3)

Hence, Option 4 is the correct option.

Given,

f(x) = px + q

As f(−1) = 1,

⇒ p(−1) + q = 1

⇒ −p + q = 1 ...........(1)

As f(2) = 13,

⇒ p(2) + q = 13

⇒ 2p + q = 13 ...........(2)

Subtracting equation (1) from equation (2), we get :

⇒ 2p + q − (−p + q) = 13 − 1

⇒ 3p = 12

⇒ p = 4

Substituting p = 4 in equation (1), we get :

⇒ −4 + q = 1

⇒ q = 5

Hence, Option 1 is the correct option.

Given,

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

Replacing $x$ by $xy$,

$f(xy) = \sqrt{1 + x^2y^2}$

Also,

$f(x) \cdot f(y) = \sqrt{1 + x^2} \times \sqrt{1 + y^2}$

⇒ $f(x) \cdot f(y) = \sqrt{(1 + x^2)(1 + y^2)}$

⇒ $f(x) \cdot f(y) = \sqrt{1 + x^2 + y^2 + x^2y^2}$

We know that $x^2 \ge 0$ and $y^2 \ge 0$, so $x^2 + y^2 \ge 0$.

⇒ $1 + x^2y^2 \le 1 + x^2 + y^2 + x^2y^2$

⇒ $\sqrt{1 + x^2y^2} \le \sqrt{1 + x^2 + y^2 + x^2y^2}$

⇒ $f(xy) \le f(x) \cdot f(y)$

Hence, Option 3 is the correct option.

Given,

$f(x) = 6x^2 + 1$ and $g(x) = 11 - 7x$

As the functions $f$ and $g$ are equal, $f(x) = g(x)$.

⇒ $6x^2 + 1 = 11 - 7x$

⇒ $6x^2 + 7x + 1 - 11 = 0$

⇒ $6x^2 + 7x - 10 = 0$

⇒ $6x^2 + 12x - 5x - 10 = 0$

⇒ $6x(x + 2) - 5(x + 2) = 0$

⇒ $(x + 2)(6x - 5) = 0$

⇒ $x = -2$ or $x = \frac{5}{6}$

∴ The required domain is $\bigg\lbrace -2, \frac{5}{6} \bigg\rbrace$

Hence, Option 3 is the correct option.

Given,

$f(x) - 3f\left(\frac{1}{x}\right) = 2x + 3$ ...........(1)

Replacing $x$ by $\frac{1}{x}$ in equation (1),

$f\left(\frac{1}{x}\right) - 3f(x) = \frac{2}{x} + 3$ ...........(2)

Multiplying equation (2) by 3,

⇒ $3f\left(\frac{1}{x}\right) - 9f(x) = \frac{6}{x} + 9$ ...........(3)

Adding equations (1) and (3), we get :

⇒ $f(x) - 9f(x) = 2x + 3 + \frac{6}{x} + 9$

⇒ $-8f(x) = 2x + \frac{6}{x} + 12$

⇒ $f(x) = -\frac{1}{8}\left(2x + \frac{6}{x} + 12\right)$

Substituting $x = 3$,

⇒ $f(3) = -\frac{1}{8}(6 + 2 + 12)$

⇒ $f(3) = -\frac{20}{8}$

⇒ $f(3) = -\frac{5}{2}$

Hence, Option 2 is the correct option.

Given,

f(x) = | x − 2 | + 1

| x − 2 | is defined for every real value of x, so | x − 2 | + 1 is a real number for every real value of x and no value of x has to be excluded.

∴ Domain of f = R

∴ Statement I is false.

Let y = f(x), then

y = | x − 2 | + 1

We know that | x − 2 | ≥ 0.

⇒ | x − 2 | + 1 ≥ 0 + 1

⇒ y ≥ 1

∴ Range of f = [1, ∞)

∴ Statement II is true.

Hence, Option 4 is the correct option.

By definition, a relation R from a set A to a set B is a subset of A × B.

∴ Reason (R) is true.

As every relation from A to B is a subset of A × B, the number of relations from A to B is the number of subsets of A × B.

We know that a set containing n elements has 2ⁿ subsets and n(A × B) = n(A) · n(B).

∴ Number of relations from A to B = 2^{n(A) · n(B)}

∴ Assertion (A) is true.

The formula in Assertion (A) is obtained exactly by counting the subsets of A × B, which is what Reason (R) states.

Hence, Option 1 is the correct option.

We know that,

Number of relations from set A to set B = 2^{n(A) · n(B)} and not 2^{n(A) + n(B) + 1}.

∴ Reason (R) is false.

Given,

Number of relations from A to B = 64

⇒ 2^{n(A) · n(B)} = 64 = 2⁶

⇒ n(A) · n(B) = 6

Also given, n(A) − n(B) = 1.

The pairs of positive integers whose product is 6 are 1 and 6, 2 and 3, 3 and 2, 6 and 1.

Out of these, only n(A) = 3 and n(B) = 2 satisfy n(A) − n(B) = 1.

∴ Assertion (A) is true.

Hence, Option 3 is the correct option.

By the definition of the modulus function,

$$|x| = \begin{cases} x, & \text{if } x \ge 0 \\[1em] -x, & \text{if } x \lt 0 \end{cases}$$

∴ Reason (R) is true.

Given,

$f(x) = \frac{x}{|x|}$, which is defined for every real number $x$ except 0.

Using the above definition,

when $x \gt 0$, $|x| = x$, so $f(x) = \frac{x}{x} = 1$,

when $x \lt 0$, $|x| = -x$, so $f(x) = \frac{x}{-x} = -1$.

So $f(x)$ takes only the values 1 and $-1$.

∴ $R_f = {-1, 1}$

∴ Assertion (A) is true.

The range of $f$ is obtained exactly by splitting $|x|$ into its two cases, which is what Reason (R) states.

Hence, Option 1 is the correct option.

Given,

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

$f(x)$ is not defined when the denominator is zero.

⇒ $x^2 - 4x + 3 = 0$

⇒ $x^2 - 3x - x + 3 = 0$

⇒ $x(x - 3) - 1(x - 3) = 0$

⇒ $(x - 3)(x - 1) = 0$

⇒ $x = 3$ or $x = 1$

∴ $D_f = \textbf{R} - {1, 3}$

∴ Assertion (A) is true.

For the function $f(x) = |x - 1|$, we know that $|x - 1| \ge 0$ for every real number $x$, and for every $y \ge 0$ the number $x = 1 + y$ gives $|x - 1| = y$.

∴ $R_f = [0, \infty)$

∴ Reason (R) is true.

Reason (R) speaks about the range of a different function and it is of no use in finding the domain stated in Assertion (A).

Hence, Option 2 is the correct option.

Every function is a relation, but a relation need not be a function. For example, the relation {(1, 2), (1, 3)} is not a function, as the element 1 has two different images.

∴ Reason (R) is false.

Given,

A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6} and R = {(1, 2), (2, 3), (3, 4), (4, 5)}

Each of the four elements 1, 2, 3 and 4 of A occurs exactly once as a first component of the ordered pairs of R, so every element of A has one and only one image in B.

∴ R is a function.

∴ Assertion (A) is true.

Hence, Option 3 is the correct option.

For the greatest integer function $f(x) = [x]$, the value $[x]$ is by definition the greatest integer not exceeding $x$, so every value of $f(x)$ is an integer.

Also, for any integer $n$, $f(n) = [n] = n$, so every integer is attained.

∴ $R_f = \textbf{I}$

∴ Assertion (A) is true.

Now, for $f(x) = \frac{1}{\sqrt{x + 2}}$, the expression $\sqrt{x + 2}$ must be a real number and it must not be zero.

⇒ $x + 2 \gt 0$

⇒ $x \gt -2$

∴ $D_f = (-2, \infty)$ and not $[-2, \infty)$, as $x = -2$ makes the denominator zero.

∴ Reason (R) is false.

Hence, Option 3 is the correct option.

Given,

A = {1, 2, 3}, B = {4, 5} and C = {5, 6}

First we find A × B and A × C.

A × B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} and

A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6)}

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

Collecting all the elements which belong to B or C or both,

⇒ B ∪ C = {4, 5, 6}

∴ A × (B ∪ C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} ...........(1)

Collecting all the ordered pairs which belong to A × B or A × C or both,

∴ (A × B) ∪ (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} ...........(2)

From equations (1) and (2), we get

Hence, A × (B ∪ C) = (A × B) ∪ (A × C).

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

The only element common to B and C is 5.

⇒ B ∩ C = {5}

∴ A × (B ∩ C) = {(1, 5), (2, 5), (3, 5)} ...........(1)

The ordered pairs common to A × B and A × C are (1, 5), (2, 5) and (3, 5).

∴ (A × B) ∩ (A × C) = {(1, 5), (2, 5), (3, 5)} ...........(2)

From equations (1) and (2), we get

Hence, A × (B ∩ C) = (A × B) ∩ (A × C).

(iii) A × (B − C) = (A × B) − (A × C)

Removing from B the element 5 which also belongs to C,

⇒ B − C = {4}

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

Removing from A × B the ordered pairs (1, 5), (2, 5) and (3, 5) which also belong to A × C,

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

From equations (1) and (2), we get

Hence, A × (B − C) = (A × B) − (A × C).

Given,

A = {2, 4, 6, 8} and B = {0, 6, 8, 9, 10}

The elements common to A and B are 6 and 8.

⇒ A ∩ B = {6, 8}

Removing from A the elements 6 and 8 which also belong to B,

⇒ A − B = {2, 4}

∴ (A ∩ B) × (A − B) = {(6, 2), (6, 4), (8, 2), (8, 4)}

Now we pick out those ordered pairs (x, y) in which x is a multiple of y.

6 is a multiple of 2, so (6, 2) is taken,

6 is not a multiple of 4, so (6, 4) is left out,

8 is a multiple of 2, so (8, 2) is taken,

8 is a multiple of 4, so (8, 4) is taken.

Thus, the corresponding ordered pairs are (6, 2), (8, 2) and (8, 4).

Hence, the required elements are (6, 2), (8, 2) and (8, 4).

Given,

A = {6, 7, 8, 10} and B = {2, 4, 5}

R = {(a, b) : a ∈ A, b ∈ B and a is divisible by b}

Taking each element of A in turn and choosing those elements of B which divide it,

when a = 6, 6 is divisible by 2 only, so b = 2,

when a = 7, 7 is not divisible by 2, 4 or 5,

when a = 8, 8 is divisible by 2 and 4, so b = 2, 4,

when a = 10, 10 is divisible by 2 and 5, so b = 2, 5.

Thus, the corresponding ordered pairs are (6, 2), (8, 2), (8, 4), (10, 2) and (10, 5).

Hence, R = {(6, 2), (8, 2), (8, 4), (10, 2), (10, 5)}.

Given,

R = {(x, y) : x + 2y < 6, x, y ∈ N}

As x and y are natural numbers, the least value of each of them is 1.

Taking each value of y in turn,

when y = 1, x + 2 < 6 ⇒ x < 4 ⇒ x = 1, 2, 3,

when y = 2, x + 4 < 6 ⇒ x < 2 ⇒ x = 1,

when y = 3, x + 6 < 6 ⇒ x < 0, which is not possible for x ∈ N.

For all the values of y greater than 3 also, there is no natural number x satisfying the condition.

Thus, the corresponding ordered pairs are (1, 1), (1, 2), (2, 1) and (3, 1).

∴ R = {(1, 1), (1, 2), (2, 1), (3, 1)}

(i) The first components of the ordered pairs of R are 1, 1, 2 and 3, and the second components are 1, 2, 1 and 1.

Hence, domain of R = {1, 2, 3} and range of R = {1, 2}.

(ii) Hence, R = {(1, 1), (1, 2), (2, 1), (3, 1)}.

Given,

R = {(x, y) : y = x + 1 and y ∈ {0, 1, 2, 3, 4, 5}}

y = x + 1 ⇒ x = y − 1

(i) Taking each value of y in turn,

when y = 0, x = −1,

when y = 1, x = 0,

when y = 2, x = 1,

when y = 3, x = 2,

when y = 4, x = 3,

when y = 5, x = 4.

Thus, the corresponding ordered pairs are (−1, 0), (0, 1), (1, 2), (2, 3), (3, 4) and (4, 5).

Hence, R = {(−1, 0), (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)}.

(ii) We write the first components −1, 0, 1, 2, 3, 4 in one closed figure and the second components 0, 1, 2, 3, 4, 5 in another, and join x to y by an arrow whenever (x, y) ∈ R.

Hence, R is represented by the arrow diagram given above.

Given,

$f = \bigg\lbrace \left(\frac{m}{n},\ m\right) : m, n \in \textbf{Z}, n \ne 0 \bigg\rbrace$

We know that,

A relation is a function only when every element of its domain has one and only one image.

A rational number can be written as a fraction in many different ways, and the rule of $f$ takes the numerator of the fraction as the image.

Taking $m = 1$, $n = 2$, we get the ordered pair $\left(\frac{1}{2},\ 1\right) \in f$

⇒ $f\left(\frac{1}{2}\right) = 1$

Taking $m = 2$, $n = 4$, we get the ordered pair $\left(\frac{2}{4},\ 2\right) \in f$

⇒ $f\left(\frac{2}{4}\right) = 2$

But $\frac{1}{2} = \frac{2}{4}$, so the same element $\frac{1}{2}$ of $\textbf{Q}$ has two different images 1 and 2.

Hence, $\bm{f}$ is not a function from $\bm{\textbf{Q}}$ to $\bm{\textbf{Z}}$, because the element $\bm{\frac{1}{2}}$ of $\bm{\textbf{Q}}$ has two different images 1 and 2.

Given,

f(x) = x², X = {−2, −1, 0, 1, 2, 3} and Y = {0, 1, 4, 7, 9, 10}

Substituting each value of x,

f(−2) = (−2)² = 4,

f(−1) = (−1)² = 1,

f(0) = 0² = 0,

f(1) = 1² = 1,

f(2) = 2² = 4,

f(3) = 3² = 9.

Each of these images 4, 1, 0, 1, 4 and 9 belongs to Y.

Thus, the corresponding ordered pairs are (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4) and (3, 9).

∴ f = {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}

Each of the six elements of X occurs exactly once as a first component, so every element of X has one and only one image in Y.

Hence, f = {(−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} and f is a function.

Given,

g = {(1, 1), (2, 3), (3, 5), (4, 7)}

The first components 1, 2, 3 and 4 are all different, so every element of the domain has one and only one image.

∴ g is a function.

Given, g(x) = αx + β

As (1, 1) ∈ g, g(1) = 1.

⇒ α(1) + β = 1

⇒ α + β = 1 ...........(1)

As (2, 3) ∈ g, g(2) = 3.

⇒ α(2) + β = 3

⇒ 2α + β = 3 ...........(2)

Subtracting equation (1) from equation (2), we get :

⇒ 2α + β − (α + β) = 3 − 1

⇒ α = 2

Substituting α = 2 in equation (1), we get :

⇒ 2 + β = 1

⇒ β = 1 − 2 = −1

Checking with the remaining pairs, g(3) = 2(3) − 1 = 5 and g(4) = 2(4) − 1 = 7, which agree with the given ordered pairs.

Hence, g is a function and α = 2, β = −1.

Given,

$f(x) = ax^2 + bx + c$

As $f(0) = 6$,

⇒ $a(0)^2 + b(0) + c = 6$

⇒ $c = 6$

As $f(2) = 11$,

⇒ $a(2)^2 + b(2) + 6 = 11$

⇒ $4a + 2b = 11 - 6$

⇒ $4a + 2b = 5$ ...........(1)

As $f(-3) = 6$,

⇒ $a(-3)^2 + b(-3) + 6 = 6$

⇒ $9a - 3b = 0$

⇒ $3a = b$ ...........(2)

Substituting $b = 3a$ in equation (1), we get :

⇒ $4a + 2(3a) = 5$

⇒ $4a + 6a = 5$

⇒ $10a = 5$

⇒ $a = \frac{1}{2}$

Substituting $a = \frac{1}{2}$ in equation (2), we get :

⇒ $b = 3 \times \frac{1}{2} = \frac{3}{2}$

Hence, $\bm{f(x) = \frac{1}{2}x^2 + \frac{3}{2}x + 6}$.

Given,

$f(x) = 2 - 3x^2$

For every real value of $x$, $2 - 3x^2$ is a real number.

∴ $D_f = \textbf{R}$

For $R_f$, let $y = f(x) = 2 - 3x^2$ ...........(1)

We know that $x^2 \ge 0$ for every real number $x$.

⇒ $3x^2 \ge 0$

⇒ $-3x^2 \le 0$

⇒ $2 - 3x^2 \le 2$

⇒ $y \le 2$

∴ $R_f = (-\infty, 2]$

Now, $f(-2) = 2 - 3(-2)^2 = 2 - 12 = -10$

The numbers associated with $-25$ are the values of $x$ for which $f(x) = -25$.

On putting $y = -25$ in equation (1), we get :

⇒ $2 - 3x^2 = -25$

⇒ $3x^2 = 2 + 25$

⇒ $3x^2 = 27$

⇒ $x^2 = 9$

⇒ $x = \pm 3$

Hence, domain of $\bm{f = \textbf{R}}$, range of $\bm{f = (-\infty, 2]}$, $\bm{f(-2) = -10}$ and the numbers associated with $\bm{-25}$ are 3 and $\bm{-3}$.

For $\sqrt{x - 3}$ to be a real number,

⇒ $x - 3 \ge 0$

⇒ $x \ge 3$

∴ Domain $= [3, \infty)$

Let $y = \sqrt{x - 3}$.

As $x \ge 3$, $x - 3 \ge 0$

⇒ $\sqrt{x - 3} \ge 0$

⇒ $y \ge 0$

∴ Range $= [0, \infty)$

Hence, domain $\bm{= [3, \infty)}$ and range $\bm{= [0, \infty)}$.

For $\sqrt{25 - x^2}$ to be a real number,

⇒ $25 - x^2 \ge 0$

⇒ $x^2 \le 25$

⇒ $-5 \le x \le 5$

∴ Domain $= [-5, 5]$

Let $y = \sqrt{25 - x^2}$.

As $-5 \le x \le 5$, we have $0 \le x^2 \le 25$

⇒ $0 \le 25 - x^2 \le 25$

⇒ $0 \le \sqrt{25 - x^2} \le 5$

⇒ $0 \le y \le 5$

∴ Range $= [0, 5]$

Hence, domain $\bm{= [-5, 5]}$ and range $\bm{= [0, 5]}$.

$|x + 1|$ is defined for every real value of $x$, so $5 - |x + 1|$ is a real number for every real value of $x$.

∴ Domain $= \textbf{R}$

Let $y = 5 - |x + 1|$.

We know that $|x + 1| \ge 0$.

⇒ $-|x + 1| \le 0$

⇒ $5 - |x + 1| \le 5$

⇒ $y \le 5$

∴ Range $= (-\infty, 5]$

Hence, domain $\bm{= \textbf{R}}$ and range $\bm{= (-\infty, 5]}$.

Given,

$$f(x) = \begin{cases} 1 + 2x, & x \lt 0 \\[1em] 3 + 5x, & x \ge 0 \end{cases}$$

For $x \lt 0$, $f(x) = 1 + 2x$, which is a straight line.

when $x = -2$, $f(x) = 1 + 2(-2) = -3$; when $x = -1$, $f(x) = -1$.

As $x$ approaches 0 from the left, $1 + 2x$ approaches 1, but the value 1 is not taken since $x = 0$ is not included in this piece.

For $x \ge 0$, $f(x) = 3 + 5x$, which is also a straight line.

when $x = 0$, $f(x) = 3$; when $x = 1$, $f(x) = 8$.

Plotting the two pieces, we get the following graph.

From the graph,

for $x \lt 0$, $2x \lt 0$ ⇒ $1 + 2x \lt 1$, so $f(x)$ takes every value less than 1,

for $x \ge 0$, $5x \ge 0$ ⇒ $3 + 5x \ge 3$, so $f(x)$ takes every value from 3 upwards.

Hence, range of $\bm{f = (-\infty, 1) \cup [3, \infty)}$.

Given,

$f(x) = 2x + 5$ and $g(x) = x^2 - 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(f + g)(x) = (2x + 5) + (x^2 - 1)$

⇒ $(f + g)(x) = x^2 + 2x + 4$

Hence, $\bm{(f + g)(x) = x^2 + 2x + 4}$, $\bm{x \in \textbf{R}}$.

Given,

$f(x) = 2x + 5$ and $g(x) = x^2 - 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(f - g)(x) = (2x + 5) - (x^2 - 1)$

⇒ $(f - g)(x) = 2x + 5 - x^2 + 1$

⇒ $(f - g)(x) = 2x + 6 - x^2$

Hence, $\bm{(f - g)(x) = 2x + 6 - x^2}$, $\bm{x \in \textbf{R}}$.

Given,

$f(x) = 2x + 5$ and $g(x) = x^2 - 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

⇒ $(fg)(x) = (2x + 5)(x^2 - 1)$

⇒ $(fg)(x) = 2x^3 - 2x + 5x^2 - 5$

⇒ $(fg)(x) = 2x^3 + 5x^2 - 2x - 5$

Hence, $\bm{(fg)(x) = 2x^3 + 5x^2 - 2x - 5}$, $\bm{x \in \textbf{R}}$.

Given,

$f(x) = 2x + 5$ and $g(x) = x^2 - 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

Here $g(x) = x^2 - 1 = 0$ when $x = 1$ or $x = -1$, so these values must be excluded.

⇒ $\left(\frac{f}{g}\right)(x) = \frac{2x + 5}{x^2 - 1}$

Hence, $\bm{\left(\frac{f}{g}\right)(x) = \frac{2x + 5}{x^2 - 1}}$, $\bm{x \in \textbf{R}}$, $\bm{x \ne 1, -1}$.

Given,

$f(x) = 2x + 5$ and $g(x) = x^2 - 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

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

Here $f(x) = 2x + 5 = 0$ when $x = -\frac{5}{2}$, so this value must be excluded.

⇒ $\left(\frac{g}{f}\right)(x) = \frac{x^2 - 1}{2x + 5}$

Hence, $\bm{\left(\frac{g}{f}\right)(x) = \frac{x^2 - 1}{2x + 5}}$, $\bm{x \in \textbf{R}}$, $\bm{x \ne -\frac{5}{2}}$.

Given,

$f(x) = 2x + 5$ and $g(x) = x^2 - 1$

Both $f$ and $g$ are defined for every real value of $x$, so the domain of each of them is $\textbf{R}$.

$(3g + 2f^2)(x) = 3g(x) + 2[f(x)]^2$

⇒ $(3g + 2f^2)(x) = 3(x^2 - 1) + 2(2x + 5)^2$

⇒ $(3g + 2f^2)(x) = 3x^2 - 3 + 2(4x^2 + 20x + 25)$

⇒ $(3g + 2f^2)(x) = 3x^2 - 3 + 8x^2 + 40x + 50$

⇒ $(3g + 2f^2)(x) = 11x^2 + 40x + 47$

Hence, $\bm{(3g + 2f^2)(x) = 11x^2 + 40x + 47}$, $\bm{x \in \textbf{R}}$.