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Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.1 | Set 1
• Last Updated : 21 Feb, 2021

### Question 1. Define a function as a set of ordered pairs.

Solution:

Let A and B be two non-empty sets. A relation from A to B, i.e., a subset of A×B, is called a function (or a mapping) from A to B, if

(i) for each a ∈ A there exists b ∈ B such that (a, b) ∈ f

(ii) (a, b) ∈ f and (a, c) ∈ f ⇒ b = c

### Question 2. Define a function as a correspondence between two sets.

Solution:

Let A and B be two non-empty sets. Then a function ‘f’ from set A to B is a rule or method or correspondence which associates elements of set A to elements of set B such that:

(i) all elements of set A are associated to elements in set B.

(ii) an element of set A is associated to a unique element in set B.

### Question 3. What is the fundamental difference between a relation and a function? Is every relation a function?

Solution:

Let ‘f’ be a function and R be a relation defined from set X to set Y.

The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to X. This is because each element of the domain of a function must have an element associated with it, whereas this is not necessary for a relation.

In relation, one element of X might be associated with one or more elements of Y, while it must be associated with only one element of Y in a function.

Thus, not every relation is a function. However, every function is necessarily a relation.

### Question 4. Let A = {–2, –1, 0, 1, 2} and f: A → Z be a function defined by f(x) = x2 – 2x – 3. Find:

(i) range of f i.e. f (A)

(ii) pre-images of 6, –3 and 5

Solution:

Given:

A = {–2, –1, 0, 1, 2}

f : A → Z such that f(x) = x2 – 2x – 3

(i) Range of f i.e. f (A)

A is the domain of the function f. Hence, range is the set of elements f(x) for all x ∈ A.

Substituting x = –2 in f(x), we get

f(–2) = (–2)2 – 2(–2) – 3

= 4 + 4 – 3

= 5

Substituting x = –1 in f(x), we get

f(–1) = (–1)2 – 2(–1) – 3

= 1 + 2 – 3

= 0

Substituting x = 0 in f(x), we get

f(0) = (0)2 – 2(0) – 3

= 0 – 0 – 3

= – 3

Substituting x = 1 in f(x), we get

f(1) = 12 – 2(1) – 3

= 1 – 2 – 3

= – 4

Substituting x = 2 in f(x), we get

f(2) = 22 – 2(2) – 3

= 4 – 4 – 3

= –3

Thus, the range of f is {-4, -3, 0, 5}.

(ii) pre-images of 6, –3 and 5

Let x be the pre-image of 6

⇒ f(x) = 6

x2 – 2x – 3 = 6

x2 – 2x – 9 = 0

x = [-(-2) ± √ ((-2)2 – 4(1) (-9))] / 2(1)

= [2 ± √ (4+36)] / 2

= [2 ± √40] / 2

= 1 ± √10

However, 1 ± √10 ∉ A

Thus, there exists no pre-image of 6.

Now, let x be the pre-image of –3

⇒ f(x) = –3

x2 – 2x – 3 = –3

x2 – 2x = 0

x(x – 2) = 0

x = 0 or 2

Clearly, both 0 and 2 are elements of A.

Thus, 0 and 2 are the pre-images of –3.

Now, let x be the pre-image of 5

⇒ f(x) = 5

x2 – 2x – 3 = 5

x2 – 2x – 8= 0

x2 – 4x + 2x – 8= 0

x(x – 4) + 2(x – 4) = 0

(x + 2)(x – 4) = 0

x = –2 or 4

However, 4 ∉ A but –2 ∈ A

Thus, –2 is the pre-images of 5.

∴ Ø, {0, 2}, -2 are the pre-images of 6, -3, 5

### Question 5. If a function f: R → R be defined by Find: f (1), f (–1), f (0), f (2).

Solution:

Given:

Let us find f (1), f (–1), f (0) and f (2).

When x > 0, f (x) = 4x + 1

Substituting x = 1 in the above equation, we get

f (1) = 4(1) + 1

= 4 + 1

= 5

When x < 0, f(x) = 3x – 2

Substituting x = –1 in the above equation, we get

f (–1) = 3(–1) – 2

= –3 – 2

= –5

When x = 0, f(x) = 1

Substituting x = 0 in the above equation, we get

f (0) = 1

When x > 0, f(x) = 4x + 1

Substituting x = 2 in the above equation, we get

f (2) = 4(2) + 1

= 8 + 1

= 9

∴ f (1) = 5, f (–1) = –5, f (0) = 1 and f (2) = 9.

### Question 6. A function f: R → R is defined by f(x) = x2. Determine

(i) range of f

(ii) {x: f(x) = 4}

(iii) {y: f(y) = –1}

Solution:

Given:

f : R → R and f(x) = x2.

(i) range of f

Domain of f = R (set of real numbers)

We know that the square of a real number is always positive or equal to zero.

∴ range of f = R+∪ {0}

(ii) {x: f(x) = 4}

Given:

f(x) = 4

we know, x2 = 4

x2 – 4 = 0

(x – 2)(x + 2) = 0

∴ x = ± 2

∴ {x: f(x) = 4} = {–2, 2}

(iii) {y: f(y) = –1}

Given:

f(y) = –1

y2 = –1

However, the domain of f is R, and for every real number y, the value of y2 is non-negative.

Hence, there exists no real y for which y2 = –1.

∴{y: f(y) = –1} = ∅

### Question 7. Let f: R+→ R, where R+ is the set of all positive real numbers, be such that f(x) = loge x. Determine

(i) the image set of the domain of f

(ii) {x: f (x) = –2}

(iii) whether f (xy) = f (x) + f (y) holds.

Solution:

Given f: R+→ R and f(x) = loge x.

(i) the image set of the domain of f

Domain of f = R+ (set of positive real numbers)

We know the value of logarithm to the base e (natural logarithm) can take all possible real values.

∴ The image set of f = R

(ii) {x: f(x) = –2}

Given f(x) = –2

loge x = –2

∴ x = e-2 [since, logb a = c ⇒ a = bc]

∴ {x: f(x) = –2} = {e-2}

(iii) Whether f (xy) = f (x) + f (y) holds.

We have f (x) = loge x ⇒ f (y) = loge y

Now, let us consider f (xy)

F (xy) = loge (xy)

f (xy) = loge (x × y) [since, logb (a×c) = logb a + logb c]

f (xy) = loge x + loge y

f (xy) = f (x) + f (y)

∴ the equation f (xy) = f (x) + f (y) holds.

### Question 8. Write the following relations as sets of ordered pairs and find which of them are functions:

(i) {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}

(ii) {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}

(iii) {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}}

Solution:

(i) {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}

When x = 1, y = 3(1) = 3

When x = 2, y = 3(2) = 6

When x = 3, y = 3(3) = 9

∴ R = {(1, 3), (2, 6), (3, 9)}

Hence, the given relation R is a function.

(ii) {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}

When x = 1, y > 1 + 1 or y > 2 ⇒ y = {4, 6}

When x = 2, y > 2 + 1 or y > 3 ⇒ y = {4, 6}

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

Hence, the given relation R is not a function.

(iii) {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}}

When x = 0, 0 + y = 3 ⇒ y = 3

When x = 1, 1 + y = 3 ⇒ y = 2

When x = 2, 2 + y = 3 ⇒ y = 1

When x = 3, 3 + y = 3 ⇒ y = 0

∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)}

Hence, the given relation R is a function.

### Question 9. Let f: R → R and g: C → C be two functions defined as f(x) = x2 and g(x) = x2. Are they equal functions?

Solution:

Given:

f: R → R ∈ f(x) = x2 and g : R → R ∈ g(x) = x2

f is defined from R to R, the domain of f = R.

g is defined from C to C, the domain of g = C.

Two functions are equal only when the domain and codomain of both the functions are equal.

In this case, the domain of f ≠ domain of g.

∴ f and g are not equal functions.

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