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

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2.

• Relations and Functions Class 12 Ex 1.1
• Relations and Functions Class 12 Ex 1.3
• Relations and Functions Class 12 Ex 1.4

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 1
Chapter Name	Relations and Functions
Exercise	Ex 1.2
Number of Questions Solved	12
Category	NCERT Solutions

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

Question 1.
Show that the function f: R —> R defined by f (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) = if f(x₁) = f(x₂)

=> x₁ = x₂
Each x ∈ R has a unique image in codomain
=> f is one-one.
(ii) For each y belonging codomain then
or 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)
=> => 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:
∴ f is not onto.

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.

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.

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.

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 < 0
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 < 0
(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 x<0 where 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.

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.

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

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.

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.

Question 9.
Let f: N —> N be defined by
f (n) =  ,if n is odd
f (n) = ,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.

Question 10.
Let A = R-{3} and B = R-{1}. consider the function f: A -> B defined by f (x) =
Solution:
Is f one-one and onto? Justify your answer.
f: A –> B where A = R – {3}, B = R – {1} f is defined by

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
=>
f is onto
f is one-one and onto. Option (a) is correct.

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