Question

The domen range of the function y = √(x² - 4) is _____.

Answer 1

${\large\underline{\sf{Solution-}}}$

Given function is

$\rm \longmapsto\:y = \sqrt{ {x}^{2} - 4}$

We know

Domain of a function is defined as set of those real values of x for which function is well defined.

So, y is defined when

$\rm \longmapsto\: {x}^{2} - 4 \geqslant 0$

$\rm \longmapsto\: {x}^{2} - {2}^{2} \geqslant 0$

$\rm \longmapsto\: (x - 2)(x + 2) \geqslant 0$

$\rm\implies \:x \leqslant - 2 \: \: or \: \: x \geqslant 2$

$\rm\implies \:x \: \in \: ( - \infty , \: - 2] \: \cup \: [2, \: \infty )$

Now, To find the range of function

We know,

Range of a function is defined as set of those real values which is obtained by assigning the values to x.

So,

$\rm \longmapsto\:y = \sqrt{ {x}^{2} - 4}$

On squaring both sides, we get

$\rm \longmapsto\: {y}^{2} = {x}^{2} - 4$

$\rm \longmapsto\: {y}^{2} + 4= {x}^{2}$

$\rm \longmapsto\:x = \sqrt{ {y}^{2} + 4 }$

Now, x is defined when

$\rm \longmapsto\: {y}^{2} + 4 \geqslant 0 \: which \: is \: always \: true \: \forall \: y \in \: real \: number$

But,

$\rm \longmapsto\:y \geqslant 0$

Hence,

$\rm\implies \:Range \: of \: function \: = [0, \: \infty )$

So, we have

$\rm\implies \:Domain \: of \: function : x \: \in \: ( - \infty , \: - 2] \: \cup \: [2, \: \infty )$

and

$\rm\implies \:Range \: of \: function \: = [0, \: \infty )$