RD Sharma Solutions for Class 12 Maths Chapter 2 Functions – PDF Download

RD Sharma Solutions for Class 12 Maths Chapter 2 – Functions are designed to help students achieve excellent academic scores. These solutions, created by subject experts, aim to boost students' confidence in understanding the concepts and solving problems efficiently within a shorter time. The solutions offer step-by-step guidance and are perfect for clearing doubts while practicing.

RD Sharma Solutions for Class 12 Maths Chapter 2 Functions covers key topics such as the definition of functions, domain, and codomain. With four exercises, this chapter helps students grasp the essential concepts needed for the CBSE Class 12 board exams. RD Sharma Solutions for Class 12 are carefully prepared following the latest CBSE syllabus and taking into account the types of questions commonly asked in the exams.

Here are some of the critical concepts covered in this chapter:

• Classification of Functions
• Types of Functions: Constant, Identity, Modulus, Integer, Exponential, Logarithmic, Reciprocal, Square Root
• Operations on Real Functions
• Kinds of Functions: One-to-One, Onto, Many-One, In-To, Bijection
• Composition of Functions: Properties and Composition of Real Functions
• Inverse of a Function: Inverse of an Element
• Graphical Representation: Relation between the graphs of a function and its inverse

Students can easily access the RD Sharma solutions to the exercises in this chapter and use them as a valuable resource for thorough revision and effective problem-solving.

Download RD Sharma Solutions for Class 12 Maths Chapter 2 Functions PDF

RD Sharma Class 12 Maths Solutions for Chapter 2 – Functions cover all the questions from the textbook, crafted by expert Mathematics teachers at Infinity Learn. Download our free PDF of Chapter 2 – Functions RD Sharma Solutions for Class 12 to boost your performance in board exams and competitive exams.

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Q. Which of the following functions from A to B are one-one and onto?

(i) f₁ = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}

(ii) f₂ = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}

(iii) f₃ = {(a, x), (b, x), (c, z), (d, z)}; A = {a, b, c, d,}, B = {x, y, z}. 

Solution:

(i) Consider f₁ = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}

Injectivity:

f₁ (1) = 3

f₁ (2) = 5

f₁ (3) = 7

⇒ Every element of A has different images in B.

So, f₁ is one-one.

Surjectivity:

Co-domain of f₁ = {3, 5, 7}

Range of f₁ =set of images = {3, 5, 7}

⇒ Co-domain = range

So, f₁ is onto.

(ii) Consider f₂ = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}

Injectivity:

f₂ (2) = a

f₂ (3) = b

f₂ (4) = c

⇒ Every element of A has different images in B.

So, f₂ is one-one.

Surjectivity:

Co-domain of f₂ = {a, b, c}

Range of f₂ = set of images = {a, b, c}

⇒ Co-domain = range

So, f₂ is onto.

(iii) Consider f₃ = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Injectivity:

f₃ (a) = x

f₃ (b) = x

f₃ (c) = z

f₃ (d) = z

⇒ a and b have the same image x.

Also c and d have the same image z

So, f₃ is not one-one.

Surjectivity:

Co-domain of f₃ ={x, y, z}

Range of f₃ =set of images = {x, z}

So, the co-domain is not same as the range.

So, f₃ is not onto.

Q. Let A = {−1, 0, 1} and f = {(x, x²) : x ∈ A}. Show that f : A → A is neither one-one nor onto.

Solution:

Given A = {−1, 0, 1} and f = {(x, x²): x ∈ A}

Also given that, f(x) = x²

Now we have to prove that given function neither one-one or nor onto.

Injectivity:

Let x = 1

Therefore f(1) = 1²=1 and

f(-1)=(-1)²=1

⇒ 1 and -1 have the same images.

So, f is not one-one.

Surjectivity:

Co-domain of f = {-1, 0, 1}

f(1) = 1² = 1,

f(-1) = (-1)² = 1 and

f(0) = 0

⇒ Range of f = {0, 1}

So, both are not same.

Hence, f is not onto

Q. Give an example of a function 

(i) Which is one-one but not onto.

(ii) Which is not one-one but onto.

(iii) Which is neither one-one nor onto.

Solution:

(i) Let f: Z → Z given by f(x) = 3x + 2

Let us check one-one condition on f(x) = 3x + 2

Injectivity:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f (x) = f(y)

⇒ 3x + 2 =3y + 2

⇒ 3x = 3y

⇒ x = y

⇒ f(x) = f(y)

⇒ x = y

So, f is one-one.

Surjectivity:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z(domain).

Let f(x) = y

⇒ 3x + 2 = y

⇒ 3x = y – 2

⇒ x = (y – 2)/3. It may not be in the domain (Z)

Because if we take y = 3,

x = (y – 2)/3 = (3-2)/3 = 1/3 ∉ domain Z.

So, for every element in the co domain there need not be any element in the domain such that f(x) = y.

Thus, f is not onto.

(ii) Example for the function which is not one-one but onto

Let f: Z → N ∪ {0} given by f(x) = |x|

Injectivity:

Let x and y be any two elements in the domain (Z),

Such that f(x) = f(y).

⇒ |x| = |y|

⇒ x = ± y

So, different elements of domain f may give the same image.

So, f is not one-one.

Surjectivity:

Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

⇒ |x| = y

⇒ x = ± y

Which is an element in Z (domain).

So, for every element in the co-domain, there exists a pre-image in the domain.

Thus, f is onto.

(iii) Example for the function which is neither one-one nor onto.

Let f: Z → Z given by f(x) = 2x² + 1

Injectivity:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

⇒ 2x²+1 = 2y²+1

⇒ 2x² = 2y²

⇒ x² = y²

⇒ x = ± y

So, different elements of domain f may give the same image.

Thus, f is not one-one.

Surjectivity:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f (x) = y

⇒ 2x²+1=y

⇒ 2x²= y − 1

⇒ x² = (y-1)/2

⇒ x = √ ((y-1)/2) ∉ Z always.

For example, if we take, y = 4,

x = ± √ ((y-1)/2)

= ± √ ((4-1)/2)

= ± √ (3/2) ∉ Z

So, x may not be in Z (domain).

Thus, f is not onto.

Q. Prove that the function f: N → N, defined by f(x) = x² + x + 1, is one-one but not onto

Solution:

Given f: N → N, defined by f(x) = x² + x + 1

Now we have to prove that given function is one-one

Injectivity:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

⇒ x² + x + 1 = y² + y + 1

⇒ (x² – y²) + (x – y) = 0 `

⇒ (x + y) (x- y ) + (x – y ) = 0

⇒ (x – y) (x + y + 1) = 0

⇒ x – y = 0 [x + y + 1 cannot be zero because x and y are natural numbers

⇒ x = y

So, f is one-one.

Surjectivity:

When x = 1

x² + x + 1 = 1 + 1 + 1 = 3

⇒ x² + x +1 ≥ 3, for every x in N.

⇒ f(x) will not assume the values 1 and 2.

So, f is not onto.

Q. Classify the following function as injection, surjection or bijection:

(i) f: N → N given by f(x) = x²

(ii) f: Z → Z given by f(x) = x²

(iii) f: N → N given by f(x) = x³

Solution:

(i) Given f: N → N, given by f(x) = x²

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x) = f(y)

x² = y²

x = y (We do not get ± because x and y are in N that is natural numbers)

So, f is an injection.

Surjection condition:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x²= y

x = √y, which may not be in N.

For example, if y = 3,

x = √3 is not in N.

So, f is not a surjection.

Also f is not a bijection.

(ii) Given f: Z → Z, given by f(x) = x²

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x) = f(y)

x² = y²

x = ±y

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x² = y

x = ± √y which may not be in Z.

For example, if y = 3,

x = ± √ 3 is not in Z.

So, f is not a surjection.

Also f is not bijection.

(iii) Given f: N → N given by f(x) = x³

Now we have to check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x) = f(y)

x³ = y³

x = y

So, f is an injection

Surjection condition:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x³= y

x = ∛y which may not be in N.

For example, if y = 3,

X = ∛3 is not in N.

So, f is not a surjection and f is not a bijection.