We have the given function as,f(x) = \( \sqrt{x-1}\)

Clearly, the term inside the root sign must be non-negative.

So, the function is valid for all values of x ≥ 1 .

Thus, the domain of the function will be, [1, ∞) .

Now, again, for x ≥ 1, the value of the function will always be greater than or equal to zero.

So, the range of the function is, [0, ∞).

Answered by Pragya Singh | 1 year agoLet R = {(a, b) : a, b, ϵ N and a < b}.Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.

Let A = (1, 2, 3} and B = {4} How many relations can be defined from A to B.

Let R = {(x, x^{2}) : x is a prime number less than 10}.

**(i) **Write R in roster form.

**(ii)** Find dom (R) and range (R).

If A = {5} and B = {5, 6}, write down all possible subsets of A × B.