[tex] {\large\underline{\sf{Solution-}}}[/tex]
Given function is
[tex]\rm \longmapsto\:y = \sqrt{ {x}^{2} - 4} [/tex]
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
[tex]\rm \longmapsto\: {x}^{2} - 4 \geqslant 0[/tex]
[tex]\rm \longmapsto\: {x}^{2} - {2}^{2} \geqslant 0[/tex]
[tex]\rm \longmapsto\: (x - 2)(x + 2) \geqslant 0[/tex]
[tex]\rm\implies \:x \leqslant - 2 \: \: or \: \: x \geqslant 2[/tex]
[tex]\rm\implies \:x \: \in \: ( - \infty , \: - 2] \: \cup \: [2, \: \infty )[/tex]
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,
[tex]\rm \longmapsto\:y = \sqrt{ {x}^{2} - 4} [/tex]
On squaring both sides, we get
[tex]\rm \longmapsto\: {y}^{2} = {x}^{2} - 4[/tex]
[tex]\rm \longmapsto\: {y}^{2} + 4= {x}^{2}[/tex]
[tex]\rm \longmapsto\:x = \sqrt{ {y}^{2} + 4 } [/tex]
Now, x is defined when
[tex]\rm \longmapsto\: {y}^{2} + 4 \geqslant 0 \: which \: is \: always \: true \: \forall \: y \in \: real \: number[/tex]
But,
[tex]\rm \longmapsto\:y \geqslant 0[/tex]
Hence,
[tex]\rm\implies \:Range \: of \: function \: = [0, \: \infty )[/tex]
So, we have
[tex]\rm\implies \:Domain \: of \: function : x \: \in \: ( - \infty , \: - 2] \: \cup \: [2, \: \infty )[/tex]
and
[tex]\rm\implies \:Range \: of \: function \: = [0, \: \infty )[/tex]
