Question

Hyy guys..
Please solve this problem... Its urgent!!! ​

Answer 1

Step-by-step explanation:

Domain

For f(x) to exist function inside root should be positive . but we know

$3 + {x}^{2} \geqslant 0$

So

$x \in \: \: r$

Now for range

let

$y = \sqrt{3 + {x}^{2} }$

Since root is always positive so y can't be negative i.e , squaring both sides

${y}^{2} =3 + {x}^{2} \\ \\ {x}^{2} = {y}^{2} - 3 \\ \\ |x| = \sqrt{ {y}^{2} - 3 }$

Which is inverse function of f(x) , so domain of this function will be range of f(x) , and for domain

${y}^{2} - 3 \geqslant 0 \\ \\ y \geqslant \sqrt{3} \: \: or \: \: y \leqslant - \sqrt{3}$

Since y is positive

$y \geqslant \sqrt{3}$

Note :

If you just need answer then you can easily notice that

$y = \sqrt{ {x}^{2} + 3 } \geqslant \sqrt{ {0}^{2} + 3} \geqslant \sqrt{3}$

[ minimum value of square of positive number is zero ]