# Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.1 | Set 1

### 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 AÃ—B, is called a function (or a mapping) from A to B, if

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

(ii) (a, b) âˆˆ f and (a, c) âˆˆ f â‡’ 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 â€˜fâ€™ 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 â€˜fâ€™ 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 = {â€“2, â€“1, 0, 1, 2} and f: A â†’ Z be a function defined by f(x) = x2 â€“ 2x â€“ 3. Find:

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

(ii) pre-images of 6, â€“3 and 5

Solution:

Given:

A = {â€“2, â€“1, 0, 1, 2}

f : A â†’ Z such that f(x) = x2 â€“ 2x â€“ 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 âˆˆ A.

Substituting x = â€“2 in f(x), we get

f(â€“2) = (â€“2)2 â€“ 2(â€“2) â€“ 3

= 4 + 4 â€“ 3

= 5

Substituting x = â€“1 in f(x), we get

f(â€“1) = (â€“1)2 â€“ 2(â€“1) â€“ 3

= 1 + 2 â€“ 3

= 0

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

f(0) = (0)2 â€“ 2(0) â€“ 3

= 0 â€“ 0 â€“ 3

= â€“ 3

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

f(1) = 12 â€“ 2(1) â€“ 3

= 1 â€“ 2 â€“ 3

= â€“ 4

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

f(2) = 22 â€“ 2(2) â€“ 3

= 4 â€“ 4 â€“ 3

= â€“3

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

(ii) pre-images of 6, â€“3 and 5

Let x be the pre-image of 6

â‡’ f(x) = 6

x2 â€“ 2x â€“ 3 = 6

x2 â€“ 2x â€“ 9 = 0

x = [-(-2) Â± âˆš ((-2)2 â€“ 4(1) (-9))] / 2(1)

= [2 Â± âˆš (4+36)] / 2

= [2 Â± âˆš40] / 2

= 1 Â± âˆš10

However, 1 Â± âˆš10 âˆ‰ A

Thus, there exists no pre-image of 6.

Now, let x be the pre-image of â€“3

â‡’ f(x) = â€“3

x2 â€“ 2x â€“ 3 = â€“3

x2 â€“ 2x = 0

x(x â€“ 2) = 0

x = 0 or 2

Clearly, both 0 and 2 are elements of A.

Thus, 0 and 2 are the pre-images of â€“3.

Now, let x be the pre-image of 5

â‡’ f(x) = 5

x2 â€“ 2x â€“ 3 = 5

x2 â€“ 2x â€“ 8= 0

x2 â€“ 4x + 2x â€“ 8= 0

x(x â€“ 4) + 2(x â€“ 4) = 0

(x + 2)(x â€“ 4) = 0

x = â€“2 or 4

However, 4 âˆ‰ A but â€“2 âˆˆ A

Thus, â€“2 is the pre-images of 5.

âˆ´ Ã˜, {0, 2}, -2 are the pre-images of 6, -3, 5

### Question 5. If a function f: R â†’ R be defined by

Find: f (1), f (â€“1), f (0), f (2).

Solution:

Given:

Let us find f (1), f (â€“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 â€“ 2

Substituting x = â€“1 in the above equation, we get

f (â€“1) = 3(â€“1) â€“ 2

= â€“3 â€“ 2

= â€“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

âˆ´ f (1) = 5, f (â€“1) = â€“5, f (0) = 1 and f (2) = 9.

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

(i) range of f

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

(iii) {y: f(y) = â€“1}

Solution:

Given:

f : R â†’ 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.

âˆ´ range of f = R+âˆª {0}

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

Given:

f(x) = 4

we know, x2 = 4

x2 â€“ 4 = 0

(x â€“ 2)(x + 2) = 0

âˆ´ x = Â± 2

âˆ´ {x: f(x) = 4} = {â€“2, 2}

(iii) {y: f(y) = â€“1}

Given:

f(y) = â€“1

y2 = â€“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 = â€“1.

âˆ´{y: f(y) = â€“1} = âˆ…

### Question 7. Let f: R+â†’ 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) = â€“2}

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

Solution:

Given f: R+â†’ 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.

âˆ´ The image set of f = R

(ii) {x: f(x) = â€“2}

Given f(x) = â€“2

loge x = â€“2

âˆ´ x = e-2 [since, logb a = c â‡’ a = bc]

âˆ´ {x: f(x) = â€“2} = {e-2}

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

We have f (x) = loge x â‡’ f (y) = loge y

Now, let us consider f (xy)

F (xy) = loge (xy)

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

f (xy) = loge x + loge y

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

âˆ´ 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 âˆˆ {1, 2, 3}, y âˆˆ {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 âˆˆ {0, 1, 2, 3}}

Solution:

(i) {(x, y): y = 3x, x âˆˆ {1, 2, 3}, y âˆˆ {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

âˆ´ 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 â‡’ y = {4, 6}

When x = 2, y > 2 + 1 or y > 3 â‡’ y = {4, 6}

âˆ´ 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 âˆˆ {0, 1, 2, 3}}

When x = 0, 0 + y = 3 â‡’ y = 3

When x = 1, 1 + y = 3 â‡’ y = 2

When x = 2, 2 + y = 3 â‡’ y = 1

When x = 3, 3 + y = 3 â‡’ y = 0

âˆ´ R = {(0, 3), (1, 2), (2, 1), (3, 0)}

Hence, the given relation R is a function.

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

Solution:

Given:

f: R â†’ R âˆˆ f(x) = x2 and g : R â†’ R âˆˆ 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 â‰  domain of g.

âˆ´ f and g are not equal functions.

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