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The relation f is defined as

It is observed that for

0 ≤ x < 3, f(x) = x²

3 < x ≤ 10, f(x) = 3x

Also, at x = 3, f(x) = 3² = 9 or f(x) = 3 × 3 = 9

i.e., at x = 3, f(x) = 9

Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.

Thus, the given relation is a function.

The relation g is defined as

It can be observed that for x = 2, g(x) = 2² = 4 and g(x) = 3 × 2 = 6

Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.

The given function is.

It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.

Hence, the domain of f is R – {2, 6}.

The given real function is.

It can be seen that is defined for (x – 1) ≥ 0.

i.e., is defined for x ≥ 1.

Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of f = [1,).

As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒

Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of f = [0,).

The given real function is f (x) = |x – 1|.

It is clear that |x – 1| is defined for all real numbers.

∴Domain of f = R

Also, for x ∈ R, |x – 1| assumes all real numbers.

Hence, the range of f is the set of all non-negative real numbers.

The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.

[Denominator is greater numerator]

Thus, range of f = [0, 1)

f, g: R → R is defined as f(x) = x + 1, g(x) = 2x – 3

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

∴(f + g) (x) = 3x – 2

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

∴ (f – g) (x) = –x + 4

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

f(x) = ax + b

(1, 1) ∈ f

⇒ f(1) = 1

⇒ a × 1 + b = 1

⇒ a + b = 1

(0, –1) ∈ f

⇒ f(0) = –1

⇒ a × 0 + b = –1

⇒ b = –1

On substituting b = –1 in a + b = 1, we obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2.

Thus, the respective values of a and b are 2 and –1.

R = {(a, b): a, b ∈ N and a = b²}

(i) It can be seen that 2 ∈ N;however, 2 ≠ 2² = 4.

Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3².

Now, 3 ≠ 9² = 81; therefore, (3, 9) ∉ N

Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4² and 4 = 2².

Now, 16 ≠ 2² = 4; therefore, (16, 2) ∉ N

Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

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

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

It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

It is observed that f is a subset of A × B.

Thus, f is a relation from A to B.

(ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.

When 1 value of y gets mapped from 2 values of X makes f a non function 

Let easy be the way when a+ b = 0 , a = -b 

ab = - a×a  which is multiple 

E.g. - 4 = -2× 2     -2 + 2 = 0

         -9 = -3×3     -3 + 3 = 0 

Hence this is many to one which is not a function