NCERT Solutions
Class 12 Maths
Relations and Functions

Ex. 1.2 Q2
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
(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
So, f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.
So, 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
So, f is not injective.
Now, −2 ∈ Z. But, there does not exist any element x ∈ Z such that
f(x) = −2 or x2 = −2
So, 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
So, f is not injective.
Now, −2 ∈ R. But, there does not exist any element x ∈ R such that
f(x) = −2 or x2 = −2
So, 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
So, f is injective.
Now, 2 ∈ N. But, there does not exist any element x ∈ N such that
f(x) = 2 or x3 = 2
So, 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
So, f is injective.
Now, 2 ∈ Z. But, there does not exist any element x ∈ Z such that
f(x) = 2 or x3 = 2
So, f is not surjective.
Hence, function f is injective but not surjective.