Chapter 1 Relations and Functions Ex 1.2 Solutions
Question - 1 : - Showthat the function f: R —> R defined by f (x) = 
isone-one onto, where R is the set of all non-zero real numbers. Is the resulttrue, if the domain R is replaced by N with co-domain being same as R ?
Answer - 1 : - (a) We observe the following properties of f:
(i) f(x) = 
if f(x1) = f(x2)
=> x1 = x2
Each x ∈ R has a unique imagein codomain
=> f is one-one.
(ii) For each y belonging codomain then
or 
thereis a unique pre image of y. => f is onto.
(b) When domain R isreplaced 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 : - Checkthe injectivity and surjectivity of the following functions:
Answer - 2 : - (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 indomain 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 itscodomain 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 todomain R of f.
=> f is not onto i.e. f is not 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 : - Provethat the Greatest Integer Function f: R->R given by f (x)=[x], is neitherone-one nor onto, where [x] denotes the greatest integer less than or equal tox.
Answer - 3 : -
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 imagesin codomain. But no fraction proper or improper belonging to codomain of f hasany pre-image in its domain.
=> f is not onto.
Question - 4 : - Showthat the Modulus Function f: R -> R given by f (x) = |x|, is neither one-onenor onto, where |x| is x, if x is positive or 0 and |x| is- x, if xis negative.
Answer - 4 : -
f: R → R is given by,
It is seen that
.
∴f(−1) = f(1), but −1 ≠ 1.
∴ f is not one-one.
Now, consider −1 ∈ R.
It is known that f(x) = 
is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = = −1.
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.
Question - 5 : - Showthat 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.
Answer - 5 : -
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 itsdomain.
∴ f is not onto.
=> f is neither into nor onto.
Question - 6 : - LetA= {1,2,3}, B = {4,5,6,7} and let f = {(1,4), (2,5), (3,6)} be a function fromA to B. Show that f is one-one.
Answer - 6 : - 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 : - Ineach of die following cases, state whether the function is one-one, onto orbijective. Justify your answer.
(i) f: R–>R defined by f(x) = 3 – 4x
(ii) f: R–>R defined by f(x) = 1 + x²
Answer - 7 : -
(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 itsdomain => 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 thesame image 2.
=> f is not one-one.
(b) No negative number belonging to its codomain has its pre-image in itsdomain
=> f is not onto. Thus f is neither one- one nor onto.
Question - 8 : - LetA and B be sets. Show that f:A x B –>B x A such that f (a, b) = (b, a) isbijective function.
Answer - 8 : - 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 itsdomain as (q, p) f is onto. Thus, f is one-one and onto i.e. it is bijective.
Question - 9 : - Letf: 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. Justifyyour answer.
Answer - 9 : -
f:N —> N, defined by
The elements 1, 2 belonging to domain of f have the same image 1 in itscodomain
=> f is not one-one.
∴ it is not injective,
(b) Every member of codomain has pre-image in its domain e.g. 1 has twopre-images 1 and 2
=> f is onto. Thus f is not one-one but it is onto
=> f is not bijective.
Question - 10 : - LetA = R-{3} and B = R-{1}. consider the function f: A -> B defined by f (x)= 
Answer - 10 : - Is f one-one and onto? Justify your answer.
f: A –> B where A = R – {3}, B = R – {1} f is defined by
