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Check the injectivity and surjectivity of the following functions: (i) f : N → N given by f(x) = x^2 (ii) f : Z → Z given by f(x) = x^2 (iii) f : R → R given by f(x) = x^2 (iv) f : N → N given by f(x) = x^3 (v) f : Z → Z given by f(x) = x^3

Class 12, NCERT Chapter 1,  Exercise 1.2, Q2

(i) f: N → N is given by,
f(x) = x2It 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) = x2It 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) = x2It 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) = x3It 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) = x3It 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.
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