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For the sake of understanding, we consider a non-negative number "2" equated to modulus of independent variable "x" like :
$$\left|x\right|=2$$
Then, the values of “x” satisfying this equation is :
$$\Rightarrow x=\pm 2$$
It is intuitive to note that values of "x" satisfying above equation is actually the intersection of modulus function "y=[x]" and "y=2" plots as shown in the figure.
Further, it is easy to realize that equating a modulus function to a negative number is meaningless. The modulus expression "|x|" always evaluates to a non-negative number for all real values of "x". Observe in the figure above that line "y=-2" does not intersect modulus plot at all.
We express these results in general form, using an expression f(x) in place of "x" as :
$$\left|\mathrm{f(x)}\right|=a;\phantom{\rule{1em}{0ex}}a>0\phantom{\rule{1em}{0ex}}\Rightarrow \mathrm{f(x)}=\pm a$$
$$\left|\mathrm{f(x)}\right|=a;\phantom{\rule{1em}{0ex}}a=0\phantom{\rule{1em}{0ex}}\Rightarrow \mathrm{f(x)}=0$$
$$\left|\mathrm{f(x)}\right|=a;\phantom{\rule{1em}{0ex}}a<0\phantom{\rule{1em}{0ex}}\Rightarrow \text{There is no solution of this equality}$$
The modulus of an expression “x-a” is interpreted to represent “distance” between “x” and “a” on the real number line. For example :
$$|x-2|=5$$
This means that the variable “x” is at a distance “5” from “2”. We see here that the values of “x” satisfying this equation is :
$$\Rightarrow x-2=\pm 5$$
Either
$$\Rightarrow x=2+5=7$$
or,
$$\Rightarrow x=2-5=-3$$
The “x = 7” is indeed at a distance “5” from “2” and “x=-3” is indeed at a distance “5” from “2”. Similarly, modulus |x|= 3 represents distance on either side of origin.
Interpretation of inequality involving modulus depends on the nature of number being compared with modulus.
Case 1 : a>0
The values of x that satisfy "less than (<)" inequality lies between intervals defined between -a and a, excluding end points of the interval. On the other hand, values of x that satisfy "greater than (>)" inequality lies in two disjointed intervals.
$$\left|x\right|<a;\phantom{\rule{1em}{0ex}}a>0\phantom{\rule{1em}{0ex}}\Rightarrow -a<x<a$$
$$\left|x\right|>a;\phantom{\rule{1em}{0ex}}a>0\phantom{\rule{1em}{0ex}}\Rightarrow x<-a\phantom{\rule{1em}{0ex}}\mathrm{or}\phantom{\rule{1em}{0ex}}x>a\phantom{\rule{1em}{0ex}}\Rightarrow x\in \left(\mathrm{-\infty}\mathrm{,a}\right)\cup \left(\mathrm{a,}\infty \right)$$
Important aspect of these inequalities is that they can be used to express intervals in compact form. For example, range of cosecant trigonometric function is $x\in \left(\mathrm{-\infty}\mathrm{,-1}\right]\cup \left[\mathrm{1,}\infty \right\}$ . Equivalently, we can write this interval as |x|≥1.
We extend these results to an expression as :
$$\left|\mathrm{f(x)}\right|<a;\phantom{\rule{1em}{0ex}}a>0\phantom{\rule{1em}{0ex}}\Rightarrow -a<\mathrm{f(x)}<a$$
For inequality involving greater than comparison with a positive number represents union of two separate intervals.
$$\left|\mathrm{f(x)}\right|>a;\phantom{\rule{1em}{0ex}}a>0\phantom{\rule{1em}{0ex}}\Rightarrow \mathrm{f(x)}<-a\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{f(x)}>a$$
Case 2 : a<0
Modulus can not be equated to negative number as modulus always evaluates to non-negative number. Clearly, modulus of an expression or variable can not be less than a negative number. However, modulus function is always greater than negative number. Hence, we conclude that :
$$\left|x\right|<a;\phantom{\rule{1em}{0ex}}a<0\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\text{There is no solution of this inequality}\phantom{\rule{1em}{0ex}}$$
$$\left|x\right|>a;\phantom{\rule{1em}{0ex}}a<0\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\text{This inequality is valid for all real values of x}\phantom{\rule{1em}{0ex}}$$
Extending these results to expression, we have :
$$\left|\mathrm{f(x)}\right|<a;\phantom{\rule{1em}{0ex}}a<0\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\text{There is no solution of this inequality}\phantom{\rule{1em}{0ex}}$$
$$\left|\mathrm{f(x)}\right|>a;\phantom{\rule{1em}{0ex}}a<0\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\text{This inequality is valid for all real values of f(x)}\phantom{\rule{1em}{0ex}}$$
Problem : Find the domain of the function given by :
$$f\left(x\right)=\frac{x}{\sqrt{\left(\left|x\right|-x\right)}}$$
Solution : The function is in rational form. The domain of the function in the numerator is "R". We are, now, required to find the value of “x” for which denominator is real and not equal to zero. Now, expression within square root is a non-negative. However, as the function is in denominator, it should not evaluate to zero either. It means that the expression within square root is positive :
$$\Rightarrow \left|x\right|-x>0$$
$$\Rightarrow \left|x\right|>x$$
If “x” is a non-negative number, then by definition, "|x| = x". This result, however, is contradictory to the inequality given above. Hence, “x” can not be non-negative. When "x" is a negative number, then the inequality holds as modulus of a real variable is always greater than negative number. It means that "x" is a negative number :
$$\Rightarrow x<0$$
Thus, domain of the given function is equal to intersection of "R" and interval "x<0". It is given by :
$$\text{Domain}=\left(-\infty ,0\right)$$
Here, we enumerate some more properties of modulus of a real variable i.e. modulus of a real number.
Let x,y and z be real variables. Then :
$$|-x|=\left|x\right|$$ $$|x-y|=0\phantom{\rule{1em}{0ex}}\iff x=y$$ $$|x+y|\le \left|x\right|+\left|y\right|$$ $$|x-y|\ge \left|\right|x|-|y\left|\right|$$ $$\left|xy\right|=\left|x\right|X\left|y\right|$$ $$\left|\frac{x}{y}\right|=\frac{\left|x\right|}{\left|y\right|};\phantom{\rule{1em}{0ex}}\left|y\right|\ne 0$$
Modulus |x-y| represents distance of x from y. Also, we know that sum of two sides of a triangle is greater than third side. Combining these two facts, we write a general property for modulus involving real numbers as :
$$|x-y|<|x-z|+|z-y|$$
There is striking similarity between modulus and square function. Both functions evaluate to non-negative values.
$$y=\left|x\right|;\phantom{\rule{1em}{0ex}}y\ge 0$$
$$y={x}^{2};\phantom{\rule{1em}{0ex}}y\ge 0$$
Their plots are similar. Besides, they behave almost alike to equalities and inequalities. We shall not discuss each of the cases as done for the modulus function, but with a specific number (4 or -4). We shall enumerate each of the possibilities, which can be easily understood in the background of discussion for modulus function.
1: Equality
$${x}^{2}=4\phantom{\rule{1em}{0ex}}\Rightarrow x=\pm 2$$
$${x}^{2}=-4\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\text{No solution}$$
2: Inequality with non-negative number
A. Less than or less than equal to
$${x}^{2}<4\phantom{\rule{1em}{0ex}}\Rightarrow -2<x<2$$
B. Greater than or greater than equal to
$${x}^{2}>4\phantom{\rule{1em}{0ex}}\Rightarrow x<-2\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x>2\phantom{\rule{1em}{0ex}}\Rightarrow (-\infty ,-2)\phantom{\rule{1em}{0ex}}\cup \phantom{\rule{1em}{0ex}}\left(\mathrm{2,}\infty \right)$$
3: Inequality with negative number
A. Less than or less than equal to
$${x}^{2}<-4\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\text{No solution}$$
B. Greater than or greater than equal to
$${x}^{2}>-4\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\text{Always true}$$
Author wishes to thank Mr. Ritesh Shah for making suggestion to remove error in the module.
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