# RD Sharma Solutions Class 11 Maths Chapter 3 – Functions (Updated For 2021-22) RD Sharma Solutions Class 11 Maths Chapter 3 – Functions: Though the chapter is always small when it comes to Functions it doesn’t mean that you can leave it or take it for granted. Understand all the concepts nicely and you will be able to solve any question that appears in your examination with the help of RD Sharma Solutions Class 11 Maths Chapter 3.

## Download RD Sharma Solutions Class 11 Maths Chapter 3 Functions PDF

RD Sharma Solutions Class 11 Maths Chapter 3

## Exercise-Wise RD Sharma Solutions Class 11 Maths Chapter 3 Functions PDF

### RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.1

The exercise-3.1 in RD Sharma Solution is one of the most basic introductory exercises of chapter 3-Functions. The exercise starts with an introduction of functions, description of functions, further explaining the meaning of domain, co-domain, and the range of functions with the help of some illustrations (examples). Exercise solutions also cover the concept of equal functions.

Go through these basic terminologies. Try to understand them with the help of your teachers or your friends and then move towards the exercise.

In this exercise, you will encounter questions like these- A function f: R -> R is defined by f(x) =. Determine the range off.

### RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.2

Though the chapters in RD Sharma Solutions are small and exercises are way too small but it doesn’t mean that you can skip any exercise. Every concept is inter-linked to others. Make sure to have at least basic knowledge about it in case a question appears in the examination.

The exercise-3.2 is very small covering only a small topic of real functions or real-valued functions.

Real-Valued Function: A function is said to be a real-valued function if a function f: A -> B if B is a subset of R (set of all real numbers).

### RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.3

The exercise-3.3 is all about finding the domain of real functions. RD Sharma Solutions provide you with a detailed step-by-step for every question you have in the RD Sharma Class 11 Mathematics.

The topic covers all the possible questions related to finding a domain of real functions. Practice questions carefully and take the help of the solutions booklet in case you feel stuck.

### RD Sharma Solutions Class 11 Maths Chapter 3 Exercise-3.4

Exercise-3.4 is the last exercise of this chapter. You have covered a large portion and now you can solve questions of exercise-3.4. This exercise a small concept of modulus and the properties of modulus function.

## Access RD Sharma Solutions Class 11 Maths Chapter 3

### 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

### 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.

### 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.

### 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

### 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.

### 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} = ∅

### 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) = logx.

(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

logx = –2

∴ x = e-2 [since, loga = c ⇒ a = bc]

∴ {x: f(x) = –2} = {e–2}

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

We have f (x) = logx ⇒ f (y) = logy

Now, let us consider f (xy)

F (xy) = log(xy)

f (xy) = log(x × y) [since, log(a×c) = loga + logc]

f (xy) = logx + logy

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

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

### 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.

### 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.

### 1. If f (x) = x2 – 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).

Solution:

Given:

f(x) = x2 – 3x + 4.

Let us find x satisfying f (x) = f (2x + 1).

We have,

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

= (2x) 2 + 2(2x) (1) + 12 – 6x – 3 + 4

= 4x2 + 4x + 1 – 6x + 1

= 4x2 – 2x + 2

Now, f (x) = f (2x + 1)

x2 – 3x + 4 = 4x2 – 2x + 2

4x2 – 2x + 2 – x2 + 3x – 4 = 0

3x2 + x – 2 = 0

3x2 + 3x – 2x – 2 = 0

3x(x + 1) – 2(x + 1) = 0

(x + 1)(3x – 2) = 0

x + 1 = 0 or 3x – 2 = 0

x = –1 or 3x = 2

x = –1 or 2/3

∴ The values of x are –1 and 2/3.

### 2. If f (x) = (x – a)2 (x – b)2, find f (a + b).

Solution:

Given:

F (x) = (x – a)2(x – b)2

Let us find f (a + b).

We have,

f (a + b) = (a + b – a)(a + b – b)2

f (a + b) = (b)2 (a)2

∴ f (a + b) = a2b2

### 3. If y = f (x) = (ax – b) / (bx – a), show that x = f (y).

Solution:

Given:

y = f (x) = (ax – b) / (bx – a) ⇒ f (y) = (ay – b) / (by – a)

Let us prove that x = f (y).

We have,

y = (ax – b) / (bx – a)

By cross-multiplying,

y(bx – a) = ax – b

bxy – ay = ax – b

bxy – ax = ay – b

x(by – a) = ay – b

x = (ay – b) / (by – a) = f (y)

∴ x = f (y)

Hence proved.

### 4. If f (x) = 1 / (1 – x), show that f [f {f (x)}] = x.

Solution:

Given:

f (x) = 1 / (1 – x)

Let us prove that f [f {f (x)}] = x.

Firstly, let us solve for f {f (x)}.

f {f (x)} = f {1/(1 – x)}

= 1 / 1 – (1/(1 – x))

= 1 / [(1 – x – 1)/(1 – x)]

= 1 / (-x/(1 – x))

= (1 – x) / -x

= (x – 1) / x

∴ f {f (x)} = (x – 1) / x

Now, we shall solve for f [f {f (x)}]

f [f {f (x)}] = f [(x-1)/x]

= 1 / [1 – (x-1)/x]

= 1 / [(x – (x-1))/x]

= 1 / [(x – x + 1)/x]

= 1 / (1/x)

∴ f [f {f (x)}] = x

Hence proved.

### 5. If f (x) = (x + 1) / (x – 1), show that f [f (x)] = x.

Solution:

Given:

f (x) = (x + 1) / (x – 1)

Let us prove that f [f (x)] = x.

f [f (x)] = f [(x+1)/(x-1)]

= [(x+1)/(x-1) + 1] / [(x+1)/(x-1) – 1]

= [[(x+1) + (x-1)]/(x-1)] / [[(x+1) – (x-1)]/(x-1)]

= [(x+1) + (x-1)] / [(x+1) – (x-1)]

= (x+1+x-1)/(x+1-x+1)

= 2x/2

= x

∴ f [f (x)] = x

Hence proved.

### 6. If ### (v) f (√-3)

Solution:

(i) f (1/2)

When, 0 ≤ x ≤ 1, f(x) = x

∴ f (1/2) = ½

(ii) f (-2)

When, x < 0, f(x) = x2

f (–2) = (–2)2

= 4

∴ f (–2) = 4

(iii) f (1)

When, x ≥ 1, f (x) = 1/x

f (1) = 1/1

∴ f(1) = 1

(iv) f (√3)

We have √3 = 1.732 > 1

When, x ≥ 1, f (x) = 1/x

∴ f (√3) = 1/√3

(v) f (√-3)

We know √-3 is not a real number and the function f(x) is defined only when x ∈ R.

∴ f (√-3) does not exist.

### (v) f (x) = (x2+2x+1)/(x2-8x+12)

Solution:

(i) f (x) = 1/x

We know, f (x) is defined for all real values of x, except for the case when x = 0.

∴ Domain of f = R – {0}

(ii) f (x) = 1/(x-7)

We know, f (x) is defined for all real values of x, except for the case when x – 7 = 0 or x = 7.

∴ Domain of f = R – {7}

(iii) f (x) = (3x-2)/(x+1)

We know, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = –1.

∴ Domain of f = R – {–1}

(iv) f (x) = (2x+1)/(x2-9)

We know, f (x) is defined for all real values of x, except for the case when x2 – 9 = 0.

x2 – 9 = 0

x2 – 32 = 0

(x + 3)(x – 3) = 0

x + 3 = 0 or x – 3 = 0

x = ± 3

∴ Domain of f = R – {–3, 3}

(v) f (x) = (x2+2x+1)/(x2-8x+12)

We know, f(x) is defined for all real values of x, except for the case when x2 – 8x + 12 = 0.

x2 – 8x + 12 = 0

x2 – 2x – 6x + 12 = 0

x(x – 2) – 6(x – 2) = 0

(x – 2)(x – 6) = 0

x – 2 = 0 or x – 6 = 0

x = 2 or 6

∴ Domain of f = R – {2, 6}

### (iv) f (x) = √(x-2)/(3-x)

Solution:

(i) f (x) = √(x-2)

We know the square of a real number is never negative.

f (x) takes real values only when x – 2 ≥ 0

x ≥ 2

∴ x ∈ [2, ∞)

∴ Domain (f) = [2, ∞)

(ii) f (x) = 1/(√(x2-1))

We know the square of a real number is never negative.

f (x) takes real values only when x2 – 1 ≥ 0

x2 – 12 ≥ 0

(x + 1) (x – 1) ≥ 0

x ≤ –1 or x ≥ 1

∴ x ∈ (–∞, –1] ∪ [1, ∞)

In addition, f (x) is also undefined when x2 – 1 = 0 because denominator will be zero and the result will be indeterminate.

x2 – 1 = 0 ⇒ x = ± 1

So, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}

x ∈ (–∞, –1) ∪ (1, ∞)

∴ Domain (f) = (–∞, –1) ∪ (1, ∞)

(iii) f (x) = √(9-x2)

We know the square of a real number is never negative.

f (x) takes real values only when 9 – x2 ≥ 0

9 ≥ x2

x2 ≤ 9

x2 – 9 ≤ 0

x2 – 32 ≤ 0

(x + 3)(x – 3) ≤ 0

x ≥ –3 and x ≤ 3

x ∈ [–3, 3]

∴ Domain (f) = [–3, 3]

(iv) f (x) = √(x-2)/(3-x)

We know the square root of a real number is never negative.

f (x) takes real values only when x – 2 and 3 – x are both positive and negative.

(a) Both x – 2 and 3 – x are positive

x – 2 ≥ 0

x ≥ 2

3 – x ≥ 0

x ≤ 3

Hence, x ≥ 2 and x ≤ 3

∴ x ∈ [2, 3]

(b) Both x – 2 and 3 – x are negative

x – 2 ≤ 0

x ≤ 2

3 – x ≤ 0

x ≥ 3

Hence, x ≤ 2 and x ≥ 3

However, the intersection of these sets is null set. Thus, this case is not possible.

Hence, x ∈ [2, 3] – {3}

x ∈ [2, 3]

∴ Domain (f) = [2, 3]

### (viii) f (x) = √(9-x2)

Solution:

(i) f (x) = (ax+b)/(bx-a)

f(x) is defined for all real values of x, except for the case when bx – a = 0 or x = a/b.

Domain (f) = R – (a/b)

Let f (x) = y

(ax+b)/(bx-a) = y

ax + b = y(bx – a)

ax + b = bxy – ay

ax – bxy = –ay – b

x(a – by) = –(ay + b)

∴ x = – (ay+b)/(a-by)

When a – by = 0 or y = a/b

Hence, f(x) cannot take the value a/b.

∴ Range (f) = R – (a/b)

(ii) f (x) = (ax-b)/(cx-d)

f(x) is defined for all real values of x, except for the case when cx – d = 0 or x = d/c. Domain (f) = R – (d/c)

Let f (x) = y

(ax-b)/(cx-d) = y

ax – b = y(cx – d)

ax – b = cxy – dy

ax – cxy = b – dy

x(a – cy) = b – dy

∴ x = (b-dy)/(a-cy)

When a – cy = 0 or y = a/c,

Hence, f(x) cannot take the value a/c.

∴ Range (f) = R – (a/c)

(iii) f (x) = √(x-1)

We know the square of a real number is never negative.

f(x) takes real values only when x – 1 ≥ 0

x ≥ 1

∴ x ∈ [1, ∞)

Thus, domain (f) = [1, ∞)

When x ≥ 1, we have x – 1 ≥ 0

Hence, √(x-1) ≥ 0 ⇒ f (x) ≥ 0

f(x) ∈ [0, ∞)

∴ Range (f) = [0, ∞)

(iv) f (x) = √(x-3)

We know the square of a real number is never negative.

f (x) takes real values only when x – 3 ≥ 0

x ≥ 3

∴ x ∈ [3, ∞)

Domain (f) = [3, ∞)

When x ≥ 3, we have x – 3 ≥ 0

Hence, √(x-3) ≥ 0 ⇒ f (x) ≥ 0

f(x) ∈ [0, ∞)

∴ Range (f) = [0, ∞)

(v) f (x) = (x-2)/(2-x)

f(x) is defined for all real values of x, except for the case when 2 – x = 0 or x = 2.

Domain (f) = R – {2}

We have, f (x) = (x-2)/(2-x)

f (x) = -(2-x)/(2-x)

= –1

When x ≠ 2, f(x) = –1

∴ Range (f) = {–1}

(vi) f (x) = |x-1|   Hence, f(x) is defined for all real numbers x.

Domain (f) = R

When, x < 1, we have x – 1 < 0 or 1 – x > 0.

|x – 1| > 0 ⇒ f(x) > 0

When, x ≥ 1, we have x – 1 ≥ 0.

|x – 1| ≥ 0 ⇒ f(x) ≥ 0

∴ f(x) ≥ 0 or f(x) ∈ [0, ∞)

Range (f) = [0, ∞)

(vii) f (x) = -|x| Now we have,  Hence, f(x) is defined for all real numbers x.

Domain (f) = R

When, x < 0, we have –|x| < 0

f (x) < 0

When, x ≥ 0, we have –x ≤ 0.

–|x| ≤ 0 ⇒ f (x) ≤ 0

∴ f (x) ≤ 0 or f (x) ∈ (–∞, 0]

Range (f) = (–∞, 0]

(viii) f (x) = √(9-x2)

We know the square of a real number is never negative.

f(x) takes real values only when 9 – x2 ≥ 0

9 ≥ x2

x2 ≤ 9

x2 – 9 ≤ 0

x2 – 32 ≤ 0

(x + 3)(x – 3) ≤ 0

x ≥ –3 and x ≤ 3

∴ x ∈ [–3, 3]

Domain (f) = [–3, 3]

When, x ∈ [–3, 3], we have 0 ≤ 9 – x2 ≤ 9

0 ≤ √(9-x2) ≤ 3 ⇒ 0 ≤ f (x) ≤ 3

∴ f(x) ∈ [0, 3]

Range (f) = [0, 3]

### (ii) f (x) = √(x-1) and g (x) = √(x+1)

Solution:

(i) f (x) = x3 + 1 and g(x) = x + 1

We have f(x): R → R and g(x): R → R

(a) f + g

We know, (f + g) (x) = f(x) + g(x)

(f + g) (x) = x3 + 1 + x + 1

= x3 + x + 2

So, (f + g) (x): R → R

∴ f + g: R → R is given by (f + g) (x) = x3 + x + 2

(b) f – g

We know, (f – g) (x) = f(x) – g(x)

(f – g) (x) = x3 + 1 – (x + 1)

= x3 + 1 – x – 1

= x3 – x

So, (f – g) (x): R → R

∴ f – g: R → R is given by (f – g) (x) = x3 – x

(c) cf (c ∈ R, c ≠ 0)

We know, (cf) (x) = c × f(x)

(cf)(x) = c(x3 + 1)

= cx3 + c

So, (cf) (x) : R → R

∴ cf: R → R is given by (cf) (x) = cx3 + c

(d) fg

We know, (fg) (x) = f(x) g(x)

(fg) (x) = (x3 + 1) (x + 1)

= (x + 1) (x2 – x + 1) (x + 1)

= (x + 1)(x2 – x + 1)

So, (fg) (x): R → R

∴ fg: R → R is given by (fg) (x) = (x + 1)2(x2 – x + 1)

(e) 1/f

We know, (1/f) (x) = 1/f (x)

1/f (x) = 1 / (x3 + 1)

Observe that 1/f(x) is undefined when f(x) = 0 or when x = – 1.

So, 1/f: R – {–1} → R is given by 1/f (x) = 1 / (x3 + 1)

(f) f/g

We know, (f/g) (x) = f(x)/g(x)

(f/g) (x) = (x3 + 1) / (x + 1)

Observe that (x3 + 1) / (x + 1) is undefined when g(x) = 0 or when x = –1.

Using x3 + 1 = (x + 1) (x2 – x + 1), we have

(f/g) (x) = [(x+1) (x2– x+1)/(x+1)]

= x2 – x + 1

∴ f/g: R – {–1} → R is given by (f/g) (x) = x2 – x + 1

(ii) f (x) = √(x-1) and g (x) = √(x+1)

We have f(x): [1, ∞) → R+ and g(x): [–1, ∞) → R+ as real square root is defined only for non-negative numbers.

(a) f + g

We know, (f + g) (x) = f(x) + g(x)

(f+g) (x) = √(x-1) + √(x+1)

Domain of (f + g) = Domain of f ∩ Domain of g

Domain of (f + g) = [1, ∞) ∩ [–1, ∞)

Domain of (f + g) = [1, ∞)

∴ f + g: [1, ∞) → R is given by (f+g) (x) = √(x-1) + √(x+1)

(b) f – g

We know, (f – g) (x) = f(x) – g(x)

(f-g) (x) = √(x-1) – √(x+1)

Domain of (f – g) = Domain of f ∩ Domain of g

Domain of (f – g) = [1, ∞) ∩ [–1, ∞)

Domain of (f – g) = [1, ∞)

∴ f – g: [1, ∞) → R is given by (f-g) (x) = √(x-1) – √(x+1)

(c) cf (c ∈ R, c ≠ 0)

We know, (cf) (x) = c × f(x)

(cf) (x) = c√(x-1)

Domain of (cf) = Domain of f

Domain of (cf) = [1, ∞)

∴ cf: [1, ∞) → R is given by (cf) (x) = c√(x-1)

(d) fg

We know, (fg) (x) = f(x) g(x)

(fg) (x) = √(x-1) √(x+1)

= √(x2 -1)

Domain of (fg) = Domain of f ∩ Domain of g

Domain of (fg) = [1, ∞) ∩ [–1, ∞)

Domain of (fg) = [1, ∞)

∴ fg: [1, ∞) → R is given by (fg) (x) = √(x2 -1)

(e) 1/f

We know, (1/f) (x) = 1/f(x)

(1/f) (x) = 1/√(x-1)

Domain of (1/f) = Domain of f

Domain of (1/f) = [1, ∞)

Observe that 1/√(x-1) is also undefined when x – 1 = 0 or x = 1.

∴ 1/f: (1, ∞) → R is given by (1/f) (x) = 1/√(x-1)

(f) f/g

We know, (f/g) (x) = f(x)/g(x)

(f/g) (x) = √(x-1)/√(x+1)

(f/g) (x) = √[(x-1)/(x+1)]

Domain of (f/g) = Domain of f ∩ Domain of g

Domain of (f/g) = [1, ∞) ∩ [–1, ∞)

Domain of (f/g) = [1, ∞)

∴ f/g: [1, ∞) → R is given by (f/g) (x) = √[(x-1)/(x+1)]

### (i) f + g(ii) f – g(iii) fg(iv) f/gFind the domain in each case.

Solution:

Given:

f(x) = 2x + 5 and g(x) = x2 + x

Both f(x) and g(x) are defined for all x ∈ R.

So, domain of f = domain of g = R

(i) f + g

We know, (f + g)(x) = f(x) + g(x)

(f + g)(x) = 2x + 5 + x2 + x

= x2 + 3x + 5

(f + g)(x) Is defined for all real numbers x.

∴ The domain of (f + g) is R

(ii) f – g

We know, (f – g)(x) = f(x) – g(x)

(f – g)(x) = 2x + 5 – (x2 + x)

= 2x + 5 – x2 – x

= 5 + x – x2

(f – g)(x) is defined for all real numbers x.

∴ The domain of (f – g) is R

(iii) fg

We know, (fg)(x) = f(x)g(x)

(fg)(x) = (2x + 5)(x2 + x)

= 2x(x2 + x) + 5(x2 + x)

= 2x3 + 2x2 + 5x2 + 5x

= 2x3 + 7x2 + 5x

(fg)(x) is defined for all real numbers x.

∴ The domain of fg is R

(iv) f/g

We know, (f/g) (x) = f(x)/g(x)

(f/g) (x) = (2x+5)/(x2+x)

(f/g) (x) is defined for all real values of x, except for the case when x2 + x = 0.

x2 + x = 0

x(x + 1) = 0

x = 0 or x + 1 = 0

x = 0 or –1

When x = 0 or –1, (f/g) (x) will be undefined as the division result will be indeterminate.

∴ The domain of f/g = R – {–1, 0}

### 3. If f(x) be defined on [–2, 2] and is given by and g(x) = f(|x|) + |f(x)|. Find g(x).

Solution:

Given: and g(x) = f(|x|) + |f(x)|. Find g(x).

Now we have, However, |x| ≥ 0 ⇒ f (|x|) = |x| – 1 when 0 < |x| ≤ 2

We also have,   ### 4. Let f, g be two real functions defined by f(x) = √(x+1) and g(x) = √(9-x2). Then, describe each of the following functions.(i) f + g(ii) g – f(iii) fg(iv) f/g(v) g/f(vi) 2f – √5g(vii) f2 + 7f(viii) 5/g

Solution:

Given:

f(x) = √(x+1) and g(x) = √(9-x2)

We know the square of a real number is never negative.

So, f(x) takes real values only when x + 1 ≥ 0

x ≥ –1, x ∈ [–1, ∞)

Domain of f = [–1, ∞)

Similarly, g(x) takes real values only when 9 – x2 ≥ 0

9 ≥ x2

x2 ≤ 9

x2 – 9 ≤ 0

x2 – 32 ≤ 0

(x + 3)(x – 3) ≤ 0

x ≥ –3 and x ≤ 3

∴ x ∈ [–3, 3]

Domain of g = [–3, 3]

(i) f + g

We know, (f + g)(x) = f(x) + g(x)

(f + g) (x) = √(x+1) + √(9-x2)

Domain of f + g = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

∴ f + g: [–1, 3] → R is given by (f + g) (x) = f(x) + g(x) = √(x+1) + √(9-x2)

(ii) g – f

We know, (g – f)(x) = g(x) – f(x)

(g – f) (x) = √(9-x2) – √(x+1)

Domain of g – f = Domain of g ∩ Domain of f

= [–3, 3] ∩ [–1, ∞)

= [–1, 3]

∴ g – f: [–1, 3] → R is given by (g – f) (x) = g(x) – f(x) = √(9-x2) – √(x+1)

(iii) fg

We know, (fg) (x) = f(x)g(x)

(fg) (x) = √(x+1) √(9-x2)

= √[(x+1) (9-x2)]

= √[x(9-x2) + (9-x2)]

= √(9x-x3+9-x2)

= √(9+9x-x2-x3)

Domain of fg = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

∴ fg: [–1, 3] → R is given by (fg) (x) = f(x) g(x) = √(x+1) √(9-x2) = √(9+9x-x2-x3)

(iv) f/g

We know, (f/g) (x) = f(x)/g(x)

(f/g) (x) = √(x+1) / √(9-x2)

= √[(x+1) / (9-x2)]

Domain of f/g = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

However, (f/g) (x) is defined for all real values of x ∈ [–1, 3], except for the case when 9 – x2 = 0 or x = ± 3

When x = ±3, (f/g) (x) will be undefined as the division result will be indeterminate.

Domain of f/g = [–1, 3] – {–3, 3}

Domain of f/g = [–1, 3)

∴ f/g: [–1, 3) → R is given by (f/g) (x) = f(x)/g(x) = √(x+1) / √(9-x2)

(v) g/f

We know, (g/f) (x) = g(x)/f(x)

(g/f) (x) = √(9-x2) / √(x+1)

= √[(9-x2) / (x+1)]

Domain of g/f = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

However, (g/f) (x) is defined for all real values of x ∈ [–1, 3], except for the case when x + 1 = 0 or x = –1

When x = –1, (g/f) (x) will be undefined as the division result will be indeterminate.

Domain of g/f = [–1, 3] – {–1}

Domain of g/f = (–1, 3]

∴ g/f: (–1, 3] → R is given by (g/f) (x) = g(x)/f(x) = √(9-x2) / √(x+1)

(vi) 2f – √5g

We know, (2f – √5g) (x) = 2f(x) – √5g(x)

(2f – √5g) (x) = 2f (x) – √5g (x)

= 2√(x+1) – √5√(9-x2)

= 2√(x+1) – √(45- 5x2)

Domain of 2f – √5g = Domain of f ∩ Domain of g

= [–1, ∞) ∩ [–3, 3]

= [–1, 3]

∴ 2f – √5g: [–1, 3] → R is given by (2f – √5g) (x) = 2f (x) – √5g (x) = 2√(x+1) – √(45- 5x2)

(vii) f2 + 7f

We know, (f2 + 7f) (x) = f2(x) + (7f)(x)

(f2 + 7f) (x) = f(x) f(x) + 7f(x)

= √(x+1) √(x+1) + 7√(x+1)

= x + 1 + 7√(x+1)

Domain of f2 + 7f is same as domain of f.

Domain of f2 + 7f = [–1, ∞)

∴ f2 + 7f: [–1, ∞) → R is given by (f2 + 7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7√(x+1)

(viii) 5/g

We know, (5/g) (x) = 5/g(x)

(5/g) (x) = 5/√(9-x2)

Domain of 5/g = Domain of g = [–3, 3]

However, (5/g) (x) is defined for all real values of x ∈ [–3, 3], except for the case when 9 – x2 = 0 or x = ± 3

When x = ±3, (5/g) (x) will be undefined as the division result will be indeterminate.

Domain of 5/g = [–3, 3] – {–3, 3}

= (–3, 3)

∴ 5/g: (–3, 3) → R is given by (5/g) (x) = 5/g(x) = 5/√(9-x2)

### Also, find (f + g) (–1), (fg) (0), (f/g) (1/2) and (g/f) (1/2).

Solution:

Given:

f(x) = log(1 – x) and g(x) = [x]

We know, f(x) takes real values only when 1 – x > 0

1 > x

x < 1, ∴ x ∈ (–∞, 1)

Domain of f = (–∞, 1)

Similarly, g(x) is defined for all real numbers x.

Domain of g = [x], x ∈ R

= R

(i) f + g

We know, (f + g) (x) = f(x) + g(x)

(f + g) (x) = log(1 – x) + [x]

Domain of f + g = Domain of f ∩ Domain of g

Domain of f + g = (–∞, 1) ∩ R

= (–∞, 1)

∴ f + g: (–∞, 1) → R is given by (f + g) (x) = log(1 – x) + [x]

(ii) fg

We know, (fg) (x) = f(x) g(x)

(fg) (x) = log(1 – x) × [x]

= [x] log(1 – x)

Domain of fg = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

∴ fg: (–∞, 1) → R is given by (fg) (x) = [x] log(1 – x)

(iii) f/g

We know, (f/g) (x) = f(x)/g(x)

(f/g) (x) = log(1 – x) / [x]

Domain of f/g = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

However, (f/g) (x) is defined for all real values of x ∈ (–∞, 1), except for the case when [x] = 0.

We have, [x] = 0 when 0 ≤ x < 1 or x ∈ [0, 1)

When 0 ≤ x < 1, (f/g) (x) will be undefined as the division result will be indeterminate.

Domain of f/g = (–∞, 1) – [0, 1)

= (–∞, 0)

∴ f/g: (–∞, 0) → R is given by (f/g) (x) = log(1 – x) / [x]

(iv) g/f

We know, (g/f) (x) = g(x)/f(x)

(g/f) (x) = [x] / log(1 – x)

However, (g/f) (x) is defined for all real values of x ∈ (–∞, 1), except for the case when log(1 – x) = 0.

log(1 – x) = 0 ⇒ 1 – x = 1 or x = 0

When x = 0, (g/f) (x) will be undefined as the division result will be indeterminate.

Domain of g/f = (–∞, 1) – {0}

= (–∞, 0) ∪ (0, 1)

∴ g/f: (–∞, 0) ∪ (0, 1) → R is given by (g/f) (x) = [x] / log(1 – x)

(a) We need to find (f + g) (–1).

We have, (f + g) (x) = log(1 – x) + [x], x ∈ (–∞, 1)

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

(f + g)(–1) = log(1 – (–1)) + [–1]

= log(1 + 1) + (–1)

= loge2 – 1

∴ (f + g) (–1) = loge2 – 1

(b) We need to find (fg) (0).

We have, (fg) (x) = [x] log(1 – x), x ∈ (–∞, 1)

Substituting x = 0 in the above equation, we get

(fg) (0) =  log(1 – 0)

= 0 × loge1

∴ (fg) (0) = 0

(c) We need to find (f/g) (1/2)

We have, (f/g) (x) = log(1 – x) / [x], x ∈ (–∞, 0)

However, 1/2 is not in the domain of f/g.

∴ (f/g) (1/2) does not exist.

(d) We need to find (g/f) (1/2)

We have, (g/f) (x) = [x] / log(1 – x), x ∈ (–∞, 0) ∪ (0, ∞)

Substituting x=1/2 in the above equation, we get

(g/f) (1/2) = [x] / log(1 – x)

= (1/2)/ log(1 – 1/2)

= 0.5/ log(1/2)

= 0 / log(1/2)

= 0

∴ (g/f) (1/2) = 0

### Important Topics from RD Sharma Solutions Class 11 Maths

Every chapter has something that you can’t afford to miss. The same is the case with Functions as well. The important topics of the chapter are:

•       Finding the domain of the function
•       Real functions
•       Finding the range of the function
•       Modulus function

## Access Other Important Chapters of RD Sharma Solutions Class 11 Maths

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