(i) Given as f: N → N, given by f(x) = x²

Let us check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in domain (N), such that f(x) = f(y).

f(x) = f(y)

x² = y²

x = y (We do not get ± because x and y are in N that is natural numbers)

So, f is an injection.

Surjection condition:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x²= y

x = √y, which may not be in N.

For example, if y = 3,

x = √3 is not in N.

So, f is not a surjection.

Also f is not a bijection.

(ii) Given f: Z → Z, given by f(x) = x²

Let us check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in domain (Z), such that f(x) = f(y).

f(x) = f(y)

x² = y²

x = ±y

Therefore, f is not an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x² = y

x = ± √y which may not be in Z.

For example, if y = 3,

x = ± √3 is not in Z.

So, f is not a surjection.

Also f is not bijection.

(iii) Given f: N → N given by f(x) = x³

Let us check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in domain (N), such that f(x) = f(y).

f(x) = f(y)

x³ = y³

x = y

So, f is an injection 

Surjection condition:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x³= y

x = ³√y which may not be in N.

For example, if y = 3,

X = ³√3 is not in N.

So, f is not a surjection and f is not a bijection.

(iv) Given f: Z → Z given by f(x) = x³

Let us check for the given function is injection, surjection and bijection condition.

Injection condition:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y)

f(x) = f(y)

x³ = y³

x = y

So, f is an injection.

Surjection condition:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x³ = y

x = ³√y which may not be in Z.

For example, if y = 3,

x = ³√3 is not in Z.

So, f is not a surjection and f is not a bijection.

(v) Given f: R → R, defined by f(x) = |x|

Let us check for the given function is injection, surjection and bijection condition.

Injection test:

Let x and y be any two elements in domain (R), such that f(x) = f(y)

f(x) = f(y)

|x| = |y|

x = ± y

Therefore, f is not an injection.

Surjection test:

Let y be any element in co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

|x| = y

x = ± y ∈ Z

Therefore, f is a surjection and f is not a bijection.