RD Sharma Solutions for Class 12 Maths – Functions

RD Sharma Solutions for Class 12 Functions provides step-by-step explanations for every exercise and example in the textbook. Each solution is written in a clear, easy-to-follow manner, helping students understand the logic behind each problem and master the topic for board and entrance exams.

Functions are a core concept in Class 12 Mathematics, forming the foundation for advanced topics like calculus, algebra, and mathematical modeling. Understanding functions is essential for mastering the CBSE Class 12 Maths syllabus, preparing for board exams, and excelling in competitive entrance tests such as JEE Main. The chapter on Functions in the RD Sharma Class 12 Maths textbook covers all important aspects, including domain, codomain, range, types of functions, and their properties.

RD Sharma Solutions for Class 12 Chapter Functions

A function is a special rule that relates each element of one set (domain) to exactly one element of another set (codomain). Functions are used to represent real-world relationships, solve equations, and analyze mathematical patterns. The RD Sharma Solutions for Class 12 introduces students to the definition of functions, how to identify domains and ranges, and how to classify different types of functions.

Download RD Sharma Class 12 Chapter Functions PDF Solutions

Students can download the RD Sharma Class 12 Maths Chapter Functions Solutions PDF, which includes all exercises, solved examples, and extra practice questions. These PDF notes are perfect for quick revision and building confidence for function-based questions in exams.

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Access Comprehensive RD Sharma Solutions for Class 12 Functions

Whether you are learning the basics of functions, working on advanced problems, or preparing for competitive exams, these RD Sharma Solutions for Functions offer a complete guide with detailed explanations and multiple solving techniques. The solutions address common doubts about domains, ranges, types of functions, inverses, and compositions, ensuring that you have a solid understanding of all key concepts.

Start your preparation now with RD Sharma Solutions for Class 12 Functions and achieve top results in your board exams and competitive tests!

Also Check: RD Sharma Solutions for Class 12 Chapter 1 Relations

RD Sharma Solutions for Functions: Step by Step Solutions

Question 1: Define a function. Give examples of different types of functions (one-one, onto, many-one, into, bijective).

Definition of Function:

A function f from set A to set B is a rule that assigns to each element in A exactly one element in B. It is denoted as f: A → B.

Types of Functions:

1. One-One (Injective) Function:

A function f: A → B is one-one if different elements in A have different images in B.

If f(x₁) = f(x₂) implies x₁ = x₂

Example: f(x) = 2x from R to R

2. Onto (Surjective) Function:

A function f: A → B is onto if every element in B is the image of at least one element in A.

Range = Codomain

Example: f(x) = x³ from R to R

3. Many-One Function:

A function f: A → B is many-one if two or more elements in A have the same image in B.

Example: f(x) = x² from R to R

4. Into Function:

A function f: A → B is into if there exists at least one element in B which is not the image of any element in A.

Range ⊂ Codomain

Example: f(x) = x² from R to R (negative numbers in R are not in range)

5. Bijective Function:

A function that is both one-one and onto.

Example: f(x) = x from R to R

Question 2: Given f: R → R by f(x) = x², determine the domain, codomain, and range. Is this function one-one or onto? Justify.

Step 1: Find Domain

Domain = R (all real numbers)

Since x² is defined for all real values of x.

Step 2: Find Codomain

Codomain = R (given in the definition)

Step 3: Find Range

Range = [0, ∞) = {y ∈ R : y ≥ 0}

Since x² ≥ 0 for all real x, and every non-negative real number can be obtained as x² for some x.

Step 4: Check if One-One

The function is NOT one-one.

Counter-example: f(2) = 4 and f(-2) = 4

Since f(2) = f(-2) but 2 ≠ -2, the function is many-one.

Step 5: Check if Onto

The function is NOT onto.

Since Range = [0, ∞) ≠ Codomain = R

For example, -1 ∈ R but there is no x ∈ R such that f(x) = -1.

Question 3: Let f: R → R be defined by f(x) = eˣ. Show that f is one-one but not onto. What happens if the codomain is changed to R⁺?

Step 1: Prove f is One-One

Let f(x₁) = f(x₂)

⟹ eˣ¹ = eˣ²

⟹ x₁ = x₂ (since exponential function is strictly increasing)

Therefore, f is one-one.

Step 2: Prove f is Not Onto

Range of f = (0, ∞) = R⁺

Codomain = R

Since Range ≠ Codomain, f is not onto.

For example, -1 ∈ R but there is no x ∈ R such that eˣ = -1.

Step 3: When Codomain is Changed to R⁺

If f: R → R⁺ where f(x) = eˣ

Then Range = R⁺ = Codomain

Therefore, f becomes onto (surjective).

Since f is already one-one, it becomes bijective.

Question 4: If f(x) = 2x + 3 and g(x) = x² + 5, find (g ∘ f)(x) and (f ∘ g)(x).

Step 1: Find (g ∘ f)(x)

(g ∘ f)(x) = g(f(x))

= g(2x + 3)

= (2x + 3)² + 5

= 4x² + 12x + 9 + 5

= 4x² + 12x + 14

Step 2: Find (f ∘ g)(x)

(f ∘ g)(x) = f(g(x))

= f(x² + 5)

= 2(x² + 5) + 3

= 2x² + 10 + 3

= 2x² + 13

Answer:

(g ∘ f)(x) = 4x² + 12x + 14

(f ∘ g)(x) = 2x² + 13

Question 5: Given f: R → R by f(x) = |x|, determine whether f is one-one, onto, or both. Explain.

Step 1: Check if One-One

The function is NOT one-one.

Counter-example: f(3) = |3| = 3 and f(-3) = |-3| = 3

Since f(3) = f(-3) but 3 ≠ -3, the function is many-one.

Step 2: Check if Onto

The function is NOT onto.

Range of f = [0, ∞) = {y ∈ R : y ≥ 0}

Codomain = R

Since Range ≠ Codomain, f is not onto.

For example, -5 ∈ R but there is no x ∈ R such that |x| = -5.

Conclusion:

f(x) = |x| is neither one-one nor onto. It is a many-one into function.

Question 6: Let f: R → R be defined by f(x) = x³. Is f invertible? If yes, find its inverse.

Step 1: Check if f is One-One

Let f(x₁) = f(x₂)

⟹ x₁³ = x₂³

⟹ x₁ = x₂ (since cube root is unique)

Therefore, f is one-one.

Step 2: Check if f is Onto

For any y ∈ R, we need to find x ∈ R such that f(x) = y

x³ = y

x = ∛y

Since cube root exists for all real numbers, f is onto.

Step 3: Conclusion on Invertibility

Since f is both one-one and onto, f is bijective.

Therefore, f is invertible.

Step 4: Find the Inverse

Let y = f(x) = x³

To find inverse, solve for x in terms of y:

x = ∛y

Therefore, f⁻¹(y) = ∛y

Or f⁻¹(x) = ∛x

Question 7: State and prove the conditions under which a function is invertible.

Statement:

A function f: A → B is invertible if and only if f is bijective (both one-one and onto).

Proof:

Part 1: If f is invertible, then f is bijective

Assume f is invertible, so f⁻¹ exists.

Proving f is one-one:

Let f(x₁) = f(x₂)

Applying f⁻¹ on both sides:

f⁻¹(f(x₁)) = f⁻¹(f(x₂))

x₁ = x₂

Therefore, f is one-one.

Proving f is onto:

For any y ∈ B, since f⁻¹ exists, there exists x = f⁻¹(y) ∈ A

such that f(x) = f(f⁻¹(y)) = y

Therefore, f is onto.

Part 2: If f is bijective, then f is invertible

Assume f is bijective (both one-one and onto).

Since f is onto, for each y ∈ B, there exists at least one x ∈ A such that f(x) = y.

Since f is one-one, for each y ∈ B, there exists exactly one x ∈ A such that f(x) = y.

Define g: B → A by g(y) = x where f(x) = y

Then g is well-defined and g = f⁻¹

Therefore, f is invertible.

Question 8: If f: R → R is defined by f(x) = sin x, is f one-one or onto? Give reasons.

Step 1: Check if f is One-One

The function is NOT one-one.

Counter-example: f(0) = sin(0) = 0 and f(π) = sin(π) = 0

Since f(0) = f(π) but 0 ≠ π, the function is many-one.

General counter-example: sin(x) = sin(π - x) for all x

Step 2: Check if f is Onto

The function is NOT onto.

Range of f = [-1, 1]

Codomain = R

Since Range ≠ Codomain, f is not onto.

For example, 2 ∈ R but there is no x ∈ R such that sin(x) = 2.

Conclusion:

f(x) = sin x is neither one-one nor onto. It is a many-one into function.

Question 9: Find all values of x for which the function f(x) = (2x + 3)/(x - 3) is defined. Also, find its range.

Step 1: Find Domain

f(x) = (2x + 3)/(x - 3)

The function is undefined when the denominator is zero.

x - 3 = 0

x = 3

Therefore, Domain = R - {3} = (-∞, 3) ∪ (3, ∞)

Step 2: Find Range

Let y = (2x + 3)/(x - 3)

Solve for x in terms of y:

y(x - 3) = 2x + 3

yx - 3y = 2x + 3

yx - 2x = 3y + 3

x(y - 2) = 3y + 3

x = (3y + 3)/(y - 2)

For x to be real, we need y - 2 ≠ 0

Therefore, y ≠ 2

Answer:

Domain: R - {3}

Range: R - {2}

Question 10: Give an example of a function which is one-one but not onto, and another which is onto but not one-one.

Example 1: One-One but Not Onto

f: R → R defined by f(x) = eˣ

Proof it's one-one:

If f(x₁) = f(x₂), then eˣ¹ = eˣ²

This implies x₁ = x₂ (exponential function is strictly increasing)

Proof it's not onto:

Range = (0, ∞) ≠ Codomain = R

For example, -1 ∈ R but no x exists such that eˣ = -1

Example 2: Onto but Not One-One

g: R → [0, ∞) defined by g(x) = x²

Proof it's onto:

For any y ∈ [0, ∞), we have x = ±√y ∈ R such that g(x) = y

Therefore, Range = [0, ∞) = Codomain

Proof it's not one-one:

g(2) = 4 and g(-2) = 4

Since g(2) = g(-2) but 2 ≠ -2, it's many-one

Question 11: Let f: R → R be defined by f(x) = x² + x. Is f one-one, onto, or bijective? Justify your answer.

Step 1: Check if f is One-One

Let f(x₁) = f(x₂)

x₁² + x₁ = x₂² + x₂

x₁² - x₂² + x₁ - x₂ = 0

(x₁ - x₂)(x₁ + x₂) + (x₁ - x₂) = 0

(x₁ - x₂)(x₁ + x₂ + 1) = 0

This gives us x₁ = x₂ or x₁ + x₂ + 1 = 0

If x₁ + x₂ + 1 = 0, then x₂ = -x₁ - 1

Counter-example: Let x₁ = 0, then x₂ = -1

f(0) = 0² + 0 = 0

f(-1) = (-1)² + (-1) = 1 - 1 = 0

Since f(0) = f(-1) but 0 ≠ -1, f is not one-one.

Step 2: Check if f is Onto

f(x) = x² + x = x(x + 1)

Complete the square: f(x) = (x + 1/2)² - 1/4

Since (x + 1/2)² ≥ 0 for all x ∈ R

f(x) ≥ -1/4 for all x ∈ R

Therefore, Range = [-1/4, ∞) ≠ Codomain = R

For example, -1 ∈ R but no x exists such that x² + x = -1

So f is not onto.

Conclusion:

f is neither one-one nor onto, hence not bijective.

Question 12: If f(x) = √(x - 1), find the domain and range of f.

Step 1: Find Domain

For f(x) = √(x - 1) to be defined, we need:

x - 1 ≥ 0

x ≥ 1

Therefore, Domain = [1, ∞)

Step 2: Find Range

Since x ≥ 1, we have x - 1 ≥ 0

Therefore, √(x - 1) ≥ 0

As x varies from 1 to ∞:

- When x = 1: f(1) = √(1-1) = 0

- As x → ∞: f(x) → ∞

Since square root function is continuous and increasing,

f takes all values from 0 to ∞.

Answer:

Domain: [1, ∞)

Range: [0, ∞)

Question 13: Let f: R → R be defined by f(x) = x² and g: R → R by g(x) = x + 1. Find (f ∘ g)(x) and (g ∘ f)(x).

Step 1: Find (f ∘ g)(x)

(f ∘ g)(x) = f(g(x))

= f(x + 1)

= (x + 1)²

= x² + 2x + 1

Step 2: Find (g ∘ f)(x)

(g ∘ f)(x) = g(f(x))

= g(x²)

= x² + 1

Answer:

(f ∘ g)(x) = x² + 2x + 1

(g ∘ f)(x) = x² + 1

Question 14: If f: R → R is defined by f(x) = log(x), for which values of x is f defined?

Step 1: Analyze the logarithm function

The natural logarithm function log(x) is defined only for positive real numbers.

Step 2: Find the domain

For f(x) = log(x) to be defined, we need:

x > 0

Step 3: Note about the given definition

The given definition f: R → R is incorrect because:

- The domain should be (0, ∞), not R

- log(x) is undefined for x ≤ 0

Correct definition:

f: (0, ∞) → R defined by f(x) = log(x)

Answer:

f(x) = log(x) is defined for x ∈ (0, ∞)

i.e., for all positive real numbers.

Question 15: Prove that the composition of two one-one functions is also one-one, and the composition of two onto functions is also onto.

Part 1: Composition of One-One Functions

Given:

f: A → B and g: B → C are both one-one functions

Let h = g ∘ f: A → C

To Prove: h is one-one

Proof:

Let x₁, x₂ ∈ A such that h(x₁) = h(x₂)

⟹ (g ∘ f)(x₁) = (g ∘ f)(x₂)

⟹ g(f(x₁)) = g(f(x₂))

Since g is one-one:

g(f(x₁)) = g(f(x₂)) ⟹ f(x₁) = f(x₂)

Since f is one-one:

f(x₁) = f(x₂) ⟹ x₁ = x₂

Therefore, h = g ∘ f is one-one.

Part 2: Composition of Onto Functions

Given:

f: A → B and g: B → C are both onto functions

Let h = g ∘ f: A → C

To Prove: h is onto

Proof:

Let z ∈ C (arbitrary element in codomain of h)

Since g is onto, there exists y ∈ B such that g(y) = z

Since f is onto, there exists x ∈ A such that f(x) = y

Therefore:

h(x) = (g ∘ f)(x) = g(f(x)) = g(y) = z

Since z was arbitrary, for every z ∈ C, there exists x ∈ A such that h(x) = z

Therefore, h = g ∘ f is onto.