Chapter 2: Relations And Functions Miscellaneous Exercise

The given relation f is defined as

It is seen that for 0 ≤ x < 3,

f(x) = x² and for 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 [Single image]

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

Therefore, the given relation is a function.

Now,

In the given relation g is defined as

It is seen that, for x = 2

g(x) = 2² = 4 and g(x) = 3 × 2 = 6

Thus, element 2 of the domain of the relation g corresponds to two different images, i.e., 4 and 6.

Therefore, this relation is not a function.

Given,

f(x) = x²

Hence,

Given function,

.

It is clearly seen that the function f is defined for all real numbers except at x = 6 and x = 2, as the denominator becomes zero otherwise.

Therefore, the domain of f is R – {2, 6}.

Given real function,

f(x) = √(x – 1)

Clearly, √(x – 1) is defined for (x – 1) ≥ 0.

So, the function f(x) = √(x – 1) is defined for x ≥ 1.

Thus, the domain of f is the set of all real numbers greater than or equal to 1.

Domain of f = [1, ∞).

Now,

As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √(x – 1) ≥ 0

Thus, the range of f is the set of all real numbers greater than or equal to 0.

Range of f = [0, ∞).

Given, a real function,

f (x) = |x – 1|

Clearly, the function |x – 1| is defined for all real numbers.

Hence,

Domain of f = R

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

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

Given function,

Substituting values and determining the images, we have

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.

Or,

We know that, for x ∈ R,

x² ≥ 0

Then,

x² + 1 ≥ x²

1 ≥ x² / (x² + 1)

Therefore, the range of f = [0, 1)

Given, the functions f, g: R → R is defined as

f(x) = x + 1, g(x) = 2x – 3

Now,

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

Thus, (f + g) (x) = 3x – 2

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

Thus, (f – g) (x) = –x + 4

f/g(x) = f(x)/g(x), g(x) ≠ 0, x ∈ R

f/g(x) = x + 1/ 2x – 3, 2x – 3 ≠ 0

Thus, f/g(x) = x + 1/ 2x – 3, x ≠ 3/2

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

And the function defined as, f(x) = ax + b

For (1, 1) ∈ f

We have, f(1) = 1

So, a × 1 + b = 1

a + b = 1 …. (i)

And for (0, –1) ∈ f

We have f(0) = –1

a × 0 + b = –1

b = –1

On substituting b = –1 in (i), we get

a + (–1) = 1 ⇒ a = 1 + 1 = 2

Therefore, the values of a and b are 2 and –1, respectively.

Given relation R = {(a, b): a, b ∈ N and a = b²}

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

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

(ii) Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3²

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

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

(iii) It’s clearly 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

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

Given,

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

So,

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)}

Also, 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’s clearly seen that f is a subset of A × B.

Therefore, f is a relation from A to B.

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