**(i)** Given,

f(x) = –|x|, x ∈ R

We know that,

As f(x) is defined for x ∈ R, the domain of f is R.

It is also seen that the range of f(x) = –|x| is all real numbers except positive real numbers.

Therefore, the range of f is given by (–∞, 0].

**(ii)** f(x) = \(
\sqrt{(9 – x^2)}\)

As \(
\sqrt{(9 – x^2)}\) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x^{2} ≥ 0.

So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3].

Now,

For any value of x in the range [–3, 3], the value of f(x) will lie between 0 and 3.

Therefore, the range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

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.