Question

For $f$ to be a function, the domain of $f$, if $f(x) = \dfrac{1}{{\sqrt {|x| - x} }}$is
A. $( - \infty ,0)$
B. $(0,\infty )$
C. $( - \infty ,\infty )$
D. $( - \infty ,0]$

Answer 1

Hint: Element inside the square root should be greater than 0. Moreover, for no value in the domain the denominator may get 0. Modulus is an operation which gives only positive value of the input quantity.

Complete step-by-step answer:
Before starting the question we need to define what a domain is. The domain of a function is the complete set of possible values of the independent variable.The definition can be modified as the domain is a set of values for which the function is defined.We know that element inside the square root should be greater than or equal to zero. But in the following case since the square root is in the denominator it can't be equal to zero. If the denominator becomes zero the function becomes indefinite.Therefore, x cannot be equal to 0 as it violates the definition of domain.Now, the element inside the square root must be greater than 0.
We have, $|x| - x > 0$
This inequality problem is to be solved by taking two cases. As the modulus operator only gives positive outcomes we have to consider cases for x to be positive and second x to be negative.
CASE 1 ( $x > 0$ )
In this case \[\left| x \right| = {\text{ }}x\]
$
   \Rightarrow x - x > 0 \\
   \Rightarrow 0 > 0 \\
$
 This is an absurd result.
CASE 2 ( $x < 0$)
In this case \[\left| x \right| = - x\] (because x is negative and multiplying it with -1 will make it positive)
$
   \Rightarrow - x - x > 0 \\
   \Rightarrow - 2x > 0 \\
   \Rightarrow x < 0 \\
$
Hence all value less than zero satisfies the equation .Thus x can be anything from negative infinity to zero but zero is excluded as explained above.
Thus the domain of function is $( - \infty ,0)$.
An open interval does not include its endpoints, and is indicated with parentheses. A closed interval is an interval which includes all its limit points, and is denoted with square brackets.
Since, zero and infinity is not included in the domain the brackets used are open brackets.

Note: While solving for domain after getting the answer it is necessary to double check by solving for boundary values. This is done to check whether the function is defined for that value in domain. (0, 90)- Open brackets does not include 0 and 90. [0, 90]- Closed brackets include 0 and 90.