NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2 (Relations and Functions)

Class 12 Maths Chapter 1 Excercise 1.2 Relations and Functions:- The NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 on Relations and Functions offer comprehensive answers to the exercise problems. These solutions are designed to help students prepare for the CBSE Class 12 board examination.

By thoroughly studying these solutions, students can master various question types related to Relations and Functions. Completing these practice questions will enhance your understanding and readiness for the Class 12 Maths Exercise 1.2 board exams. Get the NCERT Solutions for Class 12 Maths Chapter 1 Excercise 1.2 Relations and Functions from below.

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Class 12 Maths Chapter 1 Excercise 1.2 Relations and Functions

Question1. 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 :
Ans:  Given that, f: R∗→R∗  is defined by f(x)=1x

Consider x,y∈R∗  such that f(x)=f(y)

⇒1x=1y

⇒x=y

Thus, f  is one-one.
It is clear that for y∈R∗ , there exists x=1y∈R∗[ as y≠0]  such that
f(x)=1(1y)=y

Thus, f is onto.

Therefore, the given function f
 is one-one and onto.

Now, consider function g:N→R∗
 defined by g(x)=1x

We have, g(x1)=g(x2)⇒=1x1=1x2 
⇒x1=x2

Thus, g
 is one-one.
It is clear that g
 is not onto as for 1.2∈=R∗
, there does not exist any x
 in N
 such that 

g(x)  =11.2

Therefore, the function g
 is one-one but not onto.

Hence, function g is one-one but not onto.

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Question2. Check the injectivity and surjectivity of the following functions:

(i) f: N → N given by f(x) = x2

(ii) f: Z → Z given by f(x) = x2 

(iii) f: R → R given by f(x) = x2

(iv) f: N → N given by f(x) = x3

(v) f: Z → Z given by f(x) = x3

Solution :
(i) f: N → N is given by,

f(x) = x2

It is seen that for x, y ∈N, f(x) = f(y)

⇒ x2 = y2 ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

(ii) f: Z → Z is given by,

f(x) = x2

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) = x2 = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iii) f: R → R is given by,

f(x) = x2

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) = x2 = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iv) f: N → N given by,

f(x) = x3

It is seen that for x, y ∈N, f(x) = f(y)

⇒ x3 = y3 ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.

∴ f is not surjective

Hence, function f is injective but not surjective.

(v) f: Z → Z is given by,

f(x) = x3

It is seen that for x, y ∈ Z, f(x) = f(y)

⇒ x3 = y3 ⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

Question3. 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.

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Question4. 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 :

Ans: Here, f:R→R
 is given by f(x)=|x|={x−x if x⩾0 if x<0

It is clear that f(−1)=|−1|=1  and f(1)=|1|=1

Now, f(−1)=f(1)
, but −1≠1
Thus, f  is not one-one.
Now, consider −1∈R
It is known that f(x)=|x|  is always non-negative. 

Thus, there does not exist any element x
 in domain R
 such that f(x)=|x|=−1
Thus, f  is not onto.
Therefore, the modulus function is neither one-one nor onto.

Question5. Show that the Signum Function f: R → R , given by

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.

Question6. 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.

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Question7. 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 + x2

Solution : 
(i) Here, f: R→R
 is defined as f(x)=3−4x
.Let x1,x2∈R  such that f(x1)=f(x2)

⇒3−4x1=3−4x2

⇒−4x1=−4x2

⇒x1=x2

Thus, f  is one-one. 
For any real number (y)
 in R  such that f(3−y4)=3−4(3−y4)=y

Thus, f  is onto.
Therefore, the function f  is bijective.

(ii) Here, f:R→R  defined by f(x)=1+x2

Let x1,x2∈R  such that f(x1)=f(x2)

⇒1+(x1)2=1+(x2)2

⇒(x1)2=(x2)2

⇒x1=x2

Thus, f(x1)=f(x2)  does not imply that x1=x2.

For example, f(1)=f(−1)=2

Therefore, f is not one-one.
Consider an element −2  in co-domain R

It is seen that f(x)=1+x2  is positive for all x∈R.
Thus, there does not exist any x  in domain R  such that f(x)=−2
Therefore, f  is not onto.
Hence, the function f  is neither one-one nor onto.

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 : 
f:A×B→B×A
 is defined as f(a,b)=(b,a)

Let (a1,b1),(a2,b2)∈A \times B
 such that f(a1,b1)=f(a2,b2)

⇒(b1,a1)=(b2,a2)

⇒b1=b2
 and a1=a2

⇒(a1,b1)=(a2,b2)

Thus, f  is one-one.

Now, let (b,a)∈B A
 be any element.

Then, there exists (a,b)∈A B  such that f(a,b)=(b,a).[
By definition of f]
Thus, f  is onto.

Therefore, the function f is bijective.

Question9. Let N → N be defined by f(n)  for all State whether the function f is bijective. Justify your answer.

Solution : 

Question10. 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)

chapter 1-Relations and Functions Exercise 1.2

Hence, function f  is one-one and onto.

Question11. Let f: R → R be defined as f(x) = x4. 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) = x4.

Let ,x, y ∈ R  such that f(x) = f(y).

⇒ x4 = y4

⇒ x = ±y

∴f(x1) = f (x2)  does not imply that x1 = x2

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.

Question12. 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.

Class 12 Maths Chapter 1 Excercise 1.2 Relations and Functions FAQs

Q1. What are the main topics covered in Class 12 Maths Chapter 1 Exercise 1.2?

Ans. The exercise covers topics related to relations and functions, including types of relations, domain, range, and types of functions.

Q2. What is the difference between a relation and a function?

Ans. A relation is a set of ordered pairs, while a function is a specific type of relation in which each element in the domain is paired with exactly one element in the range.

Q3. What is the importance of inverse functions?

Ans. Inverse functions reverse the operation of the original function, mapping outputs back to their corresponding inputs. They are important in solving equations and in various applications in calculus.

Q4. What role do ordered pairs play in relations and functions?

Ans. Ordered pairs (x, y) represent the relationship between elements in the domain and range. In functions, each x (domain) maps to a unique y (range).

Q5. What are the key properties of functions to remember?

Ans. Key properties include domain, range, one-to-one, onto, inverse, and composition of functions. Understanding these properties helps in analyzing and solving function-related problems.