Question

What is the domain of the function :-
$f(x) = \dfrac{1}{ \sqrt{x - |x| } }$
Is it { } phi ?​

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{1}{ \sqrt{x - |x| } }$

We know,

Domain is defined as the set of those values of x for which f(x) is well defined.

Now, we know,

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: |x| = \begin{cases} &\sf{ - x \: \: when \: x \: < \: 0} \\ &\sf{ \: \: \: x \: \: when \: x \: \geqslant \: 0} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: - |x| = \begin{cases} &\sf{ \: \: \: x \: \: when \: x \: \leqslant \: 0} \\ &\sf{ \: - x \: \: when \: x \: > \: 0} \end{cases}\end{gathered}\end{gathered}$

Thus,

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: x - |x| = \begin{cases} &\sf{ \: x + x \: \: when \: x \: \leqslant \: 0} \\ &\sf{ \: x - x \: \: when \: x \: > \: 0} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: x - |x| = \begin{cases} &\sf{ \: 2x \: \: when \: x \: \leqslant \: 0} \\ &\sf{ \: 0 \: \: when \: x \: > \: 0} \end{cases}\end{gathered}\end{gathered}$

Now,

We know,

$\red{ \boxed{ \sf{ \: \frac{1}{ \sqrt{x} } \: is \: de\:fined \:when \: x \: > \: 0}}}$

Now,

We have,

$\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: x - |x| = \begin{cases} &\sf{0 \: \: when \: \: x \geqslant 0} \\ &\sf{ - ve \: \: when \: x \: < \: 0} \end{cases}\end{gathered}\end{gathered}$

$\bf\implies \:\dfrac{1}{ \sqrt{x - |x| } } \: is \: not \: de\:fined \: \forall \: x \: \in \: R$

$\bf\implies \:Domain \: of \: f(x) = \dfrac{1}{ \sqrt{x - |x| } } \: is \: \phi \: or \: \{ \: \}$