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NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2 FAQs
What does Exercise 1.2 in Class 12 Maths Chapter 1 cover?
Exercise 1.2 in Class 12 Maths Chapter 1 typically covers more advanced problems related to relations and functions. It may include questions on types of relations, equivalence relations, and composition of functions.
How many questions are there in Exercise 1.2 of Class 12 Maths Chapter 1?
The number of questions in Exercise 1.2 can vary based on the specific textbook or curriculum. Usually, it includes a set of problems to reinforce the understanding of advanced concepts introduced in the chapter.
What are the key topics addressed in Exercise 1.2?
Exercise 1.2 often deals with types of relations such as reflexive, symmetric, and transitive relations. It may also cover equivalence relations and the composition of functions.
Are Exercise 1.2 questions more challenging than Exercise 1.1?
Yes, Exercise 1.2 questions are generally more advanced and may require a deeper understanding of the concepts introduced in Exercise 1.1. They often involve higher-order thinking and problem-solving skills.
How can I prepare for Exercise 1.2 questions?
Review the concepts from Exercise 1.1 thoroughly. Understand the definitions and properties related to different types of relations and functions. Practice solving similar problems and seek guidance from your teacher if needed.
NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2 (Relations and Functions)
NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2: Get accurate error free NCERT Solutions for Class 12 Maths chapter 1 Relations and Functions Exercise 1.2.
Krati Saraswat12 Jan, 2024
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NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2 (Relations and Functions)
NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2 Relations and Functions is prepared by academic team of Physics Wallah. We have prepared
NCERT Solutions
for all exercise of chapter 1. Given below is step by step solutions of all questions given in NCERT textbook for Chapter 1 Relations and Functions.
NCERT Solutions For Class 12 Maths Chapter 1 Overview
NCERT Solutions for Class 12 Maths Chapter 1 contains all the important topics for the exams. Our experts created these questions for the students to ace the examination. This article contains all the important questions and their easy to understand answers for the better understanding.
Students are advised to go through these questions to clarify their concepts better. These questions are created to help students in the better understanding of the NCERT Solutions for Class 12 Maths Chapter 1.
To help students understand and practise chapter concepts, our team of experts at Physics Wallah has developed thorough solutions for NCERT Class 12 Maths, Chapter 1. The purpose of these questions is to make explanations easier to understand for students.
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NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2
Solve The Following Questions NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2 of Relations and Functions:
Question
1. Show that the function f : R
*
→ R
*
defined by f(x) = 1/x is one-one and onto, where : R
*
is the set of all non-zero real numbers. Is the result true, if the domain : R
*
is replaced by N with co-domain being same as : R
*
?
Solution :
Hence, function
g
is one-one but not onto.
Question
2. Check the injectivity and surjectivity of the following functions:
(i) f: N → N given by f(x) = x
2
(ii) f: Z → Z given by f(x) = x
2
(iii) f: R → R given by f(x) = x
2
(iv) f: N → N given by f(x) = x
3
(v) f: Z → Z given by f(x) = x
3
Solution :
(i) f: N → N is given by,
f(x) = x
2
It is seen that for x, y ∈N, f(x) = f(y)
⇒ x
2
= y
2
⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x
2
= 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
(ii) f: Z → Z is given by,
f(x) = x
2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x
2
= −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iii) f: R → R is given by,
f(x) = x
2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x
2
= −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iv) f: N → N given by,
f(x) = x
3
It is seen that for x, y ∈N, f(x) = f(y)
⇒ x
3
= y
3
⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x
3
= 2.
∴ f is not surjective
Hence, function f is injective but not surjective.
(v) f: Z → Z is given by,
f(x) = x
3
It is seen that for x, y ∈ Z, f(x) = f(y)
⇒ x
3
= y
3
⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x
3
= 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
Question
3. Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution :
f: R → R is given by,
f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
Question
4. Show that the Modulus Function f : R → R, given by f(x) = |x| is neither one-one nor onto, where is |x| if x is positive or 0 and |-x| is -x if x is negative.
Solution :
Question
5. Show that the Signum Function
f
: R → R , given by
is neither one-one nor onto.
Solution :
f
: R → R
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴ f is not one-one.
Now, as f(x) ) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R , there does not exist any x in domain R such that
f(x) = −2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
Question
6. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Solution :
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.
Question
7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f: R → R defined by f(x) = 3 − 4x
(ii) f: R → R defined by f(x) = 1 + x
2
Solution :
Question
8. Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.
Solution :
Question
9. Let N → N be defined by f(n) for all
State whether the function f is bijective. Justify your answer.
Solution :
Question
10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2/x-3) Is f one-one and onto? Justify your answer.
Solution :
A = R – {3} and B = R – {1}
f : A → B defined by f(x) = (x-2/x-3)
Let x, y ∈ A such that f(x) = f(y)
Hence, function f is one-one and onto.
Question
11. Let f: R → R be defined as f(x) = x
4
. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Solution :
f: R→ R is defined as f(x) = x
4
.
Let
,x, y ∈ R such that f(x) = f(y).
⇒ x
4
= y
4
⇒ x = ±y
∴f(x
1
) = f (
x2
) does not imply that x
1
= x
2
For instance,
f(1) = f(-1) = 1
∴ f is not one-one.
Consider an element 2 in co-domain R . It is clear that there does not exist any x in domain R such that f(x) = 2.
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is D.
Question
12. Let f: R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Solution :
f: R → R be defined as f(x) = 3x.
Let
,x, y ∈ R such that f(x) = f(y).
⇒ 3x = 3y
⇒ x = y
∴ f is one - one
Also, for any real number (y) in co-domain R, there exists y/3 in R such that f(y/3) = 3(y/3) = y
∴ f is onto.
Hence, function f is one-one and onto.
The correct answer is A.
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Hence, function
g
is one-one but not onto.
is neither one-one nor onto.
Solution :
f
: R → R
Question
9. Let N → N be defined by f(n) for all
State whether the function f is bijective. Justify your answer.
Solution :
Question
10. Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2/x-3) Is f one-one and onto? Justify your answer.
Solution :
A = R – {3} and B = R – {1}
f : A → B defined by f(x) = (x-2/x-3)
Let x, y ∈ A such that f(x) = f(y)
Hence, function f is one-one and onto.
Question
11. Let f: R → R be defined as f(x) = x
4
. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Solution :
f: R→ R is defined as f(x) = x
4
.
