Question

The domain of the function,$f\left( x \right) = \dfrac{1}{{\sqrt {\left| x \right| - x} }}$
A) $\left( { - \infty ,0} \right)$
B) $\left( { - \infty ,\infty } \right) - \left\{ 0 \right\}$
C) $\left( { - \infty ,\infty } \right)$
D) $\left( {0,\infty } \right)$

Answer 1

Hint:The value of the term in the denominator is a square root function, which can never be equal to 0 and less than 0.

Complete step-by-step solution:
The function is given as
$f\left( x \right) = \dfrac{1}{{\sqrt {\left| x \right| - x} }}$
In order to find the domain of the function , it is necessary to find the points at which the function exists or the points at which the function is defined.
The analysis of the function reveals that the term in the denominator consists of 2 functions.
Square root function: For this function the value inside the square root is always greater than or equal to 0.
Modulus function: the modulus of a function always returns a positive value.
Considering denominator of the function which is the critical part,
$\sqrt {\left| x \right| - x} $ , The term in the denominator is a square root function and this term will always be greater than 0.
$\left| x \right| - x > 0 \cdots \left( 1 \right)$
There are 2 cases in this equality present at equation (1),
Case (i) If positive values $\left( {x > 0} \right)$ are substituted in $\sqrt {\left| x \right| - x} $ .
For instance, put any positive value say $x = 1$ ,
$
  \sqrt {\left| 1 \right| - 1} \\
  \sqrt {1 - 1} \\
   = 0 \\
 $
So, for any positive values of $x$ , the value of the term in the denominator will always be 0.
But the term in the denominator can never be 0.
Therefore, positive values $x > 0$ or $\left( {0,\infty } \right)$ are not in its domain.
Case (ii) If negative values are substituted in $\sqrt {\left| x \right| - x} $.
For instance, put$x = - 2$ ,
$
   = \sqrt {\left| { - 2} \right| - \left( { - 2} \right)} \\
   = \sqrt {2 + 2} \\
   = \sqrt 4 \\
   = 2 \\
 $
For any negative value $\left( {x < 0} \right)$ the term in the denominator will never be less than or equal 0.
Therefore, negative values are in its domain.
The function exists for $x < 0$ or $\left( { - \infty ,0} \right)$ .
Thus, the domain of $f\left( x \right) = \dfrac{1}{{\sqrt {\left| x \right| - x} }}$is $\left( { - \infty ,0} \right)$

Hence, the correct option is (A).

Note:The domain is the set of all the permissible values of x which makes the function defined. If for some value the function is not defined, then that value will be in its domain.
The range is set of all the values which are the images of $x$. These are the output values of a function.
The modulus function always gives the positive values no matter what number is put into it.
For instance, $\left| { - 2} \right| = 2,\left| { - 4} \right| = 4$.