**(i)** f (\( \dfrac{1}{2}\))

When, 0 ≤ x ≤ 1, f(x) = x

f (\( \dfrac{1}{2}\)) = \( \dfrac{1}{2}\)

**(ii)** f (-2)

When, x < 0, f(x) = x^{2}

f (–2) = (–2)^{2}

= 4

f (–2) = 4

**(iii)** f (1)

When, x ≥ 1, f (x) = \( \dfrac{1}{x}\)

f (1) = \( \dfrac{1}{1}\)

f(1) = 1

**(iv)** \( f(\sqrt{3})\)

We have \( \sqrt{3}\) = 1.732 > 1

When, x ≥ 1, f (x) = \( \dfrac{1}{x}\)

\(f( \sqrt{3})\)= \(\dfrac{1}{ \sqrt{3}}\)

**(v)** \(f(- \sqrt{3})\)

We know \(- \sqrt{3}\) is not a real number and the function f(x) is defined only when x ∈ R.

\(f(- \sqrt{3})\) does not exist.

Answered by Sakshi | 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.