RD Sharma Solutions for Class 12 Maths Exercise 2.4 Chapter 2 Function (Updated for 2021-22)

RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4

RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4: RD Sharma Solutions for Class 12, Maths Chapter 2, helps students who wish to achieve a good academic score in the exam. RD Sharma Maths Solutions for Class 12 Chapter 2 are expertly designed to increase students’ confidence to understand the concepts covered in this chapter and how to solve problems in a short period of time.

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RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4

Our experts prepare these materials based on the Class 12 CBSE syllabus, keeping in mind the types of questions asked in the RD Sharma solution. Chapter 2 Functions explains the function and domains of functions and functions. It has four practices. Students can easily get answers to problems in RD Sharma Solutions for Class 12.

Access answers to RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4

Exercise 2.4 Page No: 2.68

1. State with reason whether the following functions have inverse:
(i) f: {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Solution:

(i) Given f: {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

We have:

f (1) = f (2) = f (3) = f (4) = 10

⇒ f is not one-one.

⇒ f is not a bijection.

So, f does not have an inverse.

(ii) Given g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

from the question it is clear that g (5) = g (7) = 4

⇒ f is not one-one.

⇒ f is not a bijection.

So, f does not have an inverse.

(iii) Given h: {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Here, different elements of the domain have different images in the co-domain.

⇒ h is one-one.

Also, each element in the co-domain has a pre-image in the domain.

⇒ h is onto.

⇒ h is a bijection.

Therefore h inverse exists.

⇒ h has an inverse and it is given by

h-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}

2. Find f −1 if it exists: f: A → B, where 

(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.

(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2

Solution:

(i) Given A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.

So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}

Here, different elements of the domain have different images in the co-domain.

Clearly, this is one-one.

Range of f = Range of f = B

so, f is a bijection and,

Thus, f -1 exists.

Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) Given A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2

So, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}

Here, different elements of the domain have different images in the co-domain.

Clearly, f is one-one.

But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)

⇒ f is not a bijection.

So, f -1does not exist.

3. Consider f: {1, 2, 3} → {a, b, c} and g: {a, b, c} → {apple, ball, cat} defined as f (1) = a, f (2) = b, f (3) = c, g (a) = apple, g (b) = ball and g (c) = cat. Show that f, g and gof are invertible. Find f−1, g−1 and gof−1and show that (gof)−1 = f −1o g−1

Solution:

Given f = {(1, a), (2, b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.

So, f and g are invertible. 

Now,

-1 = {(a ,1) , (b , 2) , (3,c)} and g-1 = {(apple, a), (ball , b), (cat , c)}

So, f-1 o g-1= {apple, 1), (ball, 2), (cat, 3)}……… (1)

f: {1,2,3,} → {a, b, c} and g: {a, b, c} → {apple, ball, cat}

So, gof: {1, 2, 3} → {apple, ball, cat}

⇒ (gof) (1) = g (f (1)) = g (a) = apple

(gof) (2) = g (f (2))

= g (b)

= ball,

And (gof) (3) = g (f (3))

= g (c)

= cat 

∴ gof = {(1, apple), (2, ball), (3, cat)}

Clearly, gof is a bijection.

So, gof is invertible. 

(gof)-1 = {(apple, 1), (ball, 2), (cat, 3)}……. (2)

Form (1) and (2), we get

(gof)-1 = f-1 o g -1

4. Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f: A → B, g: B → C be defined as f(x) = 2x + 1 and g(x) = x2 − 2. Express (gof)−1 and f−1 og−1 as the sets of ordered pairs and verify that (gof)−1 = f−1 og−1.

Solution:

Given that f (x) = 2x + 1

⇒ f= {(1, 2(1) + 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) + 1)}

= {(1, 3), (2, 5), (3, 7), (4, 9)}

Also given that g(x) = x2−2

⇒ g = {(3, 32−2), (5, 52−2), (7, 72−2), (9, 92−2)}

= {(3, 7), (5, 23), (7, 47), (9, 79)}

Clearly f and g are bijections and, hence, f−1: B→ A and g−1: C→ B exist.

So, f−1= {(3, 1), (5, 2), (7, 3), (9, 4)} 

And g−1= {(7, 3), (23, 5), (47, 7), (79, 9)}

Now, (f−1 o g−1): C→ A

f−1 o g−1 = {(7, 1), (23, 2), (47, 3), (79, 4)}……….(1)

Also, f: A→B and g: B → C,

⇒ gof: A → C, (gof) −1 : C→ A

So, f−1 o g−1and (gof)−1 have same domains.

(gof) (x) = g (f (x))

=g (2x + 1)

=(2x +1 )2− 2

⇒ (gof) (x) = 4x+ 4x + 1 − 2

⇒ (gof) (x) = 4x2+ 4x −1

Then, (gof) (1) = g (f (1)) 

= 4 + 4 − 1 

=7,

(gof) (2) = g (f (2))

= 4(2)2 + 4(2) – 1 = 23,

(gof) (3) = g (f (3))

= 4(3)2 + 4(3) – 1 = 47 and 

(gof) (4) = g (f (4))

= 4(4)2 + 4(4) − 1 = 79

So, gof = {(1, 7), (2, 23), (3, 47), (4, 79)}

⇒ (gof)– 1 = {(7, 1), (23, 2), (47, 3), (79, 4)}…… (2)

From (1) and (2), we get:

(gof)−1 = f−1 o g−1

5. Show that the function f: Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1

Solution:

Given function f: Q → Q, defined by f(x) = 3x + 5

Now we have to show that the given function is invertible.

Injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y)

⇒ 3x + 5 = 3y + 5

⇒ 3x = 3y

⇒ x = y

so, f is one-one.

Surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ 3x +5 = y 

⇒ 3x = y – 5

⇒ x = (y -5)/3 belongs to Q domain

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f-1:

Let f-1(x) = y…… (1)

⇒ x = f(y)

⇒ x = 3y + 5

⇒ x −5 = 3y

⇒ y = (x – 5)/3

Now substituting this value in (1) we get

So, f-1(x) = (x – 5)/3

6. Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Solution:

Given f: R → R given by f(x) = 4x + 3

Now we have to show that the given function is invertible.

Consider injection of f:

Let x and y be two elements of domain (R),

Such that f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

So, f is one-one.

Now surjection of f:

Let y be in the co-domain (R),

Such that f(x) = y.

⇒ 4x + 3 = y 

⇒ 4x = y -3

⇒ x = (y-3)/4 in R (domain)

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f -1

Let f-1(x) = y……. (1)

⇒ x = f (y)

⇒ x = 4y + 3

⇒ x − 3 = 4y

⇒ y = (x -3)/4

Now substituting this value in (1) we get

So, f-1(x) = (x-3)/4

7. Consider f: R → R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1(x) = √ (x-4) where R+ is the set of all non-negative real numbers.

Solution:

Given f: R → R+ → [4, ∞) given by f(x) = x2 + 4.

Now we have to show that f is invertible,

Consider injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y) 

⇒ x+ 4 = y+ 4

⇒ x= y2

⇒ x = y (as co-domain as R+)

So, f is one-one

Now surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ x2 + 4 = y

⇒ x2 = y – 4

⇒ x = √ (y-4) in R

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f-1:

Let f−1 (x) = y…… (1)

⇒ x = f (y)

⇒ x = y2 + 4

⇒ x − 4 = y2

⇒ y = √ (x-4)

So, f-1(x) = √ (x-4)

Now substituting this value in (1) we get,

So, f-1(x) = √ (x-4)

8. If f(x) = (4x + 3)/ (6x – 4), x ≠ (2/3) show that fof(x) = x, for all x ≠ (2/3). What is the inverse of f?

Solution:

It is given that f(x) = (4x + 3)/ (6x – 4), x ≠ 2/3

Now we have to show fof(x) = x

(fof)(x) = f (f(x))

= f ((4x+ 3)/ (6x – 4))

= (4((4x + 3)/ (6x -4)) + 3)/ (6 ((4x +3)/ (6x – 4)) – 4)

= (16x + 12 + 18x – 12)/ (24x + 18 – 24x + 16)

= (34x)/ (34)

= x

Therefore, fof(x) = x for all x ≠ 2/3

=> fof = 1

Hence, the given function f is invertible and the inverse of f is f itself.

9. Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with

f-1(x) = (√(x +6)-1)/3 

Solution:

Given f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x – 5

We have to show that f is invertible.

Injectivity of f:

Let x and y be two elements of domain (R+),

Such that f(x) = f(y)

⇒ 9x+ 6x – 5 = 9y+ 6y − 5

⇒ 9x+ 6x = 9y+ 6y

⇒ x = y (As, x, y ∈ R+)

So, f is one-one.

Surjectivity of f:

Let y is in the co domain (Q)

Such that f(x) = y

⇒ 9x2 + 6x – 5 = y

⇒ 9x2 + 6x = y + 5

⇒ 9x2 + 6x +1 = y + 6 (By adding 1 on both sides)

⇒ (3x + 1)2 = y + 6

⇒ 3x + 1 = √(y + 6)

⇒ 3x = √ (y + 6) – 1

⇒ x = (√ (y + 6)-1)/3 in R+ (domain)

f is onto.

So, f is a bijection and hence, it is invertible.

Now we have to find f-1

Let f−1(x) = y….. (1)

⇒ x = f (y)

⇒ x = 9y+ 6y − 5

⇒ x + 5 = 9y+ 6y

⇒ x + 6 = 9y2+ 6y + 1 (adding 1 on both sides)

⇒ x + 6 = (3y + 1)2

⇒ 3y + 1 = √ (x + 6)

⇒ 3y =√(x +6) -1

⇒ y = (√ (x+6)-1)/3

Now substituting this value in (1) we get,

So, f-1(x) = (√ (x+6)-1)/3 

10. If f: R → R be defined by f(x) = x3 −3, then prove that f−1 exists and find a formula for f−1. Hence, find f−1 (24) and f−1 (5).

Solution:

Given f: R → R be defined by f(x) = x3 −3

Now we have to prove that f−1 exists

Injectivity of f:

Let x and y be two elements in domain (R),

Such that, x3 − 3 = y3 − 3

⇒ x3 = y3

⇒ x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R)

Such that f(x) = y

⇒ x3 – 3 = y

⇒ x3 = y + 3 

⇒ x = ∛(y+3) in R

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f -1:

Let f-1(x) = y…….. (1)

⇒ x= f(y)

⇒ x = y− 3

⇒ x + 3 = y3

⇒ y = ∛(x + 3) = f-1(x) [from (1)]

So, f-1(x) = ∛(x + 3)

Now, f-1(24) = ∛ (24 + 3)

= ∛27

= ∛33

= 3

And f-1(5) =∛ (5 + 3)

= ∛8

= ∛23

= 2

11. A function f: R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).

Solution:

Given that f: R → R is defined as f(x) = x3 + 4

Injectivity of f:

Let x and y be two elements of domain (R),

Such that f (x) = f (y)

⇒ x3 + 4 = y3 + 4

⇒ x3 = y3

⇒ x = y

So, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R),

Such that f(x) = y.

⇒ x3 + 4 = y 

⇒ x3 = y – 4

⇒ x = ∛ (y – 4) in R (domain)

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Finding f-1:

Let f−1 (x) = y…… (1)

⇒ x = f (y)

⇒ x = y+ 4

⇒ x − 4 = y3

⇒ y =∛ (x-4)

So, f-1(x) =∛ (x-4) [from (1)]

f-1 (3) = ∛(3 – 4)

= ∛-1

= -1

RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4: Important Topics

Let us have a look at some of the important concepts that are discussed in this chapter.

  • Classification of functions
    • Types of functions
      • Constant function
      • Identity function
      • Modulus function
      • Integer function
      • Exponential function
      • Logarithmic function
      • Reciprocal function
      • Square root function
    • Operations on real functions
    • Kinds of functions
      • One-one function
      • On-to function
      • Many one function
      • In to function
      • Bijection
    • Composition of functions
    • Properties of the composition of functions
    • Composition of real function
    • Inverse of a function
    • Inverse of an element
    • Relation between graphs of a function and its inverse

We have provided complete details of RD Sharma Solutions Class 12 Maths Chapter 2 Exercise 2.4. If you have any queries related to CBSE Class 12 Exam, feel free to ask us in the comment section below.

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