Given f: Z → Z given by f(x) = x^{3}

Now we have to 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)

x^{3} = y^{3}

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

x^{3} = y

x =\( 3\sqrt{y}\) which may not be in Z.

For example, if y = 3,

x =\( 3\sqrt{3}\) is not in Z.

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

Answered by Aaryan | 1 year agoLet A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y

Express R and R^{-1} as sets of ordered pairs. Determine also

**(i) **the domain of R^{‑1}

**(ii)** The Range of R.

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\dfrac{\sqrt{(x+6)-1}}{3}\)

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