# NCERT Solutions For Class 12 Maths Chapter 1 Relations and Functions Ex 1.2

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## NCERT Solutions For Class 12 Chapter 1 Maths Relations and Functions Ex 1.2

Get Free NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Relations and Functions Exercise 1.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 1 Class 12 Relations and Functions Ex 1.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

Topics and Sub Topics in Class 11 Maths Chapter 1 Relations and Functions Ex 1.2:

• 1 – Relations and Functions
• 1.1 Introduction
• 1.2 Types of Relations
• 1.3 Types of Functions
• 1.4 Composition of Functions and Invertible Function
• 1.5 Binary Operations

### NCERT Solutions for Class 12 Maths – Chapter 1 – Relations and Functions Ex 1.2

Ex 1.2 Class 12 Maths Question 1.
Show that the function f: R —> R defined by f (x) = (\ frac { 1 }{ x } ) is one-one 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:
(a) We observe the following properties of f:
(i) f(x) = (\ frac { 1 }{ x } ) if f(x1) = f(x2)
(frac { 1 }{ { x }_{ 1 } } =frac { 1 }{ { x }_{ 2 } } )
=> x1 = x2
Each x ∈ R has a unique image in codomain
=> f is one-one.
(ii) For each y belonging codomain then
(y= frac { 1 }{ x } ) or (x= frac { 1 }{ y } ) there is a unique pre image of y.
=> f is onto.

(b) When domain R is replaced by N. codomain remaining the same, then f: N—> R If f(x1) = f(x2)
=> (frac { 1 }{ { n }_{ 1 } } =frac { 1 }{ { n }_{ 2 } } ) => n1 = n2 where n1; n2 ∈ N
=> f is one-one.
But for every real number belonging to codomain may not have a pre-image in N.
eg:(frac { 1 }{ 2 } ,frac { 3 }{ 2 } ,N)
∴ f is not onto.

Ex 1.2 Class 12 Maths Question 2.
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 given by f (x) = x²
(a) f(x1) =>f(x2)
=>x12 = x22 =>x1 = x2
∴ f is one-one i.e. it is injective.
(b) There are such member of codomain which have no image in domain N.
e.g. 3 ∈ codomain N. But there is no pre-image in domain of f.
=> f is not onto i.e. not surjective.
(ii) f: z —> z given by f(x) = x²
(a) f (-1) = f (1) = 1 => -1 and 1 have the same image.
∴ f is not one-one i.e. not injective.
(b) There are many such elements belonging to codomain have no pre-image in its codomain z.
e.g. 3 ∈ codomain z but √3 ∉ domain z of f,
∴ f is not onto i.e. not surjective
(iii) f: R->R, given by f(x) = x²
(a) f is not one-one since f(-1) = f(1) = 1
– 1 and 1 have the same image i.e., f is not injective
(b) – 2∈ codomain R off but √-2 does not belong to domain R of f.
=> f is not onto i.e. f is not surjective.
(iv) Injective but not surjective.
(v) Injective but not surjective.

Ex 1.2 Class 12 Maths 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 given by f (x) = [x]
(a) f(1. 2) = 1, f(1. 5) = 1 => f is not one-one
(b) All the images of x e R belonging to its domain have integers as the images in codomain. But no fraction proper or improper belonging to codomain of f has any pre-image in its domain.
=> f is not onto.

Ex 1.2 Class 12 Maths Question 4.
Show that the Modulus Function f: R -> R given by f (x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is- x, if xis negative.
Solution:
f: R->R given by f(x) = |x|
(a) f(-1) = |-1| = 1,f(1) = |1| = 1
=> -1 and 1 have the same image
∴ f is not one-one
(b) No negative value belonging to codomain of f has any pre-image in its domain
∴ f is not onto. Hence, f is neither one-one nor onto.

Ex 1.2 Class 12 Maths Question 5.
Show that the Signum Function f: R–>R given by
f(x) = 1, if x > 0
f(x) = 0, if x = 0
f(x) = – 1, if x
is neither one-one nor onto.
Solution:
f: R–>R given by
f(x) = 1, if x > 0
f(x) = 0, if x = 0
f(x) = -1, if x
(a) f(x1) = f(x2) = 1
∴ 1 and 2 have the same image i.e.
f(x1) = f(x2) = 1 for x>0
=> x1≠x2
Similarly f(x1) = f(x2) = – 1, for x1 ≠ x2 => f is not one-one.
(b) Except – 1,0,1 no other member of codomain of f has any pre-image in its domain.
∴ f is not onto.
=> f is neither into nor onto.

Ex 1.2 Class 12 Maths 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:
A= {1,2,3},B= {4,5,6,7} f= {(1,4), (2,5), (3,6) }.
Every member of A has a unique image in B ∴ f is one – one.

Ex 1.2 Class 12 Maths Question 7.
In each of die 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) f: R —> R defined by 3 – 4x,
f (x1) = 3 – 4x1, f(x2) = 3 – 4x2
(a) f(x1) = f(x2) =>3 – 4x1 = 3 – 4x2
=> x1 = x2. This shows that f is one-one
(b) f(x) = y = 3 – 4x
(x= frac { 3-y }{ 4 } )
For every value of y belonging to its codomain. There is a pre-image in its domain => f is onto.
Hence, f is one-one onto

(ii) f: R—>R given by f(x)= 1 + x²
(a) f(1) = 1 + 1 = 2,f(-1) = 1 +1 = 2
∴ f (-1) = f (1) = 2 i.e.-1 and 1 have the same image 2.
=> f is not one-one.
(b) No negative number belonging to its codomain has its pre-image in its domain
=> f is not onto. Thus f is neither one- one nor onto.

Ex 1.2 Class 12 Maths Question 8.
Let A and B be sets. Show that f:A x B –>B x A such that f (a, b) = (b, a) is bijective function.
Solution:
We have f: (A x B) —> B x A such that f (a, b) = b, a
(a) ∴ f(a1, b1)= (b1, a1) f(a2, b2) = (b2, a2) f(a1, b1) = f(a2, b2)
=>(b1, a1)
= (b2, a2)
=> b1 = b2 and a1 = a2 f is one-one
(b) Every member (p, q) belonging to its codomain has its pre-image in its domain as (q, p) f is onto. Thus, f is one-one and onto i.e. it is bijective.

Ex 1.2 Class 12 Maths Question 9.
Let f: N —> N be defined by
f (n) =
(frac { n+1 }{ 2 } ) ,if n is odd
f (n) = (frac { n }{ 2 } ) ,if n is even
for all n∈N

State whether the function f is bijective. Justify your answer.
Solution:
f: N —> N, defined by The elements 1, 2 belonging to domain of f have the same image 1 in its codomain
=> f is not one-one.
∴ it is not injective,
(b) Every member of codomain has pre-image in its domain e.g. 1 has two pre-images 1 and 2
=> f is onto. Thus f is not one-one but it is onto
=> f is not bijective.

Ex 1.2 Class 12 Maths Question 10.
Let A = R-{3} and B = R-{1}. consider the function f: A -> B defined by f (x) = (left( frac { x-2 }{ x-3 } right) )
Solution:
Is f one-one and onto? Justify your answer.
f: A –> B where A = R – {3}, B = R – {1} f is defined by  Ex 1.2 Class 12 Maths Question 11.
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(-1) = (-1)4 = 1,f(1) = 14 = 1
∴ – 1, 1 have the same image 1 => f is not one- one
Further – 2 in the codomain of f has no pre-image in its domain.
∴ f is not onto i.e. f is neither one-one nor onto Option (d) is correct.

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 is defined by f (x) = 3x
(a) f(x1) = 3x1, f(x2) = 3x2
=> f(x1) = f(x2)
=> 3x1 = 3x2
=> x1 = x2
=> f is one-one
(b) for every member y belonging to co-domain has pre-image x in domain of f.
∵ y = 3x
=>(x= frac { y }{ 3 } )
f is onto
f is one-one and onto. Option (a) is correct.

### NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions in Hindi Medium Ex 1.2            