Modulus or Absolut value function.
Modulus of x represented as $\vert x \vert$ is also called as absolute value of x.
Let us try to understand the absolute value of a number. for example the absolute value of numbers 3 is 3 , $\pi$ is $\pi$, and $\sqrt{13}$ is $\sqrt{13}$ itself.
Where as if we talk about the absolute value of $-3$ is also 3, that of $-\pi$ is $\pi$ and $-\sqrt{13}$ is $\sqrt{13}$.
Technically what does it mean is $\vert 3 \vert = \vert -3 \vert =3$ , $\vert \pi \vert = \vert -\pi \vert = \pi$ and that of $\vert \sqrt{13} \vert = \vert -\sqrt{13} \vert = \sqrt{13}$.
From above discussion it is very clear that modulus of any number ‘x’is the number ‘x’ itself if x is positive, where as Modulus of ‘x’ is ‘-x’ if x is negative. that is $\vert x \vert = \begin{cases} x&,if\, x\ge 0 \\ -x&,if \,x\lt 0 \end{cases}$
That means, if we have $\vert x \vert = 7$ that means $x = \pm 7$ since $\vert 7 \vert = \vert -7 \vert =7$
Definition of Modulus of x. Modulus of unknown x is defined as the positive square root of $x^2$. i.e $\vert x \vert = \sqrt{x^2}$.
Geometrical Significance of Modulus.
Geometrically Modulus of ‘x’ is the distance of point $(x,0)$ from origin $(0,0)$.
Let P be any point on real axis such that point is $P(x,0)$ then distance of point P from origin OP is given by,
$OP=\sqrt{(x-0)^2 +(0-0)^2}=\sqrt{x^2}=\vert x \vert $
Hence from above we can say that modulus is a distance of point x on real axis from origin. Since it is a distance hence cannot be negative.
Similarly we have $\vert f(x) \vert = \begin{cases} f(x)&, if\;f(x)\ge 0 \\ -f(x)&, if\;f(x) \lt 0 \end{cases}$
Ways to solve questions involving Modulus function.
Step by Step method to solve.
Based on If $\vert f(x) \vert = a$ where $a\ge 0$ then $f(x)=\pm a$
Let us see why?
we know that $f(x) =\begin{cases} f(x) &, if\;f(x) \ge 0 \\ -f(x) &, if\;f(x) \lt 0 \end{cases}$
Case I :- Let $f(x) \ge 0$ Hence $\vert f(x) \vert = f(x)$
$\Rightarrow \vert f(x) \vert = f(x) = a\;\;–(1)$
Case II :- Let $f(x) \lt 0$ Hence $\vert f(x) \vert = -f(x) $
$\Rightarrow \vert f(x) \vert = -f(x) = a$ Multiply both sides by $-1$ we get,
$\Rightarrow f(x) = -a\;\;–(2)$
From (1) and (2) we get , $f(x) = \pm a$
Note :- If $\vert f(x) \vert = a$ where $ a\lt 0$ then $x\in \emptyset$
(1) Find x such that $\vert x \vert = 2$
Sol :- If $\vert x \vert = 2$
$\Rightarrow x =\pm 2 $
(2) Find x such that $\vert \cfrac{2x}{1-x} \vert = 1$
Sol :- For $\vert \cfrac{2x}{1-x} \vert = 1 $
$\Rightarrow \cfrac{2x}{1-x} = \pm 1$
$\therefore 2x = \pm (1-x)$
Hence $2x = 1 – x \;and\;2x = -1 + x$
$\therefore x=\cfrac{1}{3} , -1$
Based on If $\vert f(x) \vert \le a$ where $a\ge 0$ then $-a\le f(x)\le a$
Let us see why?
we know that $f(x) =\begin{cases} f(x) &, if\;f(x) \ge 0 \\ -f(x) &, if\;f(x) \lt 0 \end{cases}$
Case I :- Let $f(x) \ge 0\;\forall x \in A$ (Where A is the interval in which f(x) is positive )
Hence in interval A we can replace $\vert f(x) \vert = f(x)$ in the given inequation.
Hence we will get $ \vert f(x) \vert = f(x) \le a\;\forall x\in B$ (Where interval of x is B that we get from inequality in case I).
Hence now the interval in which f(x) is positive and satisfy the given in-equation will be $x\in A\cap B$. Because interval A will ensure that f(x) is positive and interval B is the solution of in-equation, hence the x that satisfy both conditions will satisfy the in-equation.
Now let interval C is $C\in A\cap B$
Hence from case I we get $f(x)\le a\,\forall x\in C$
Case II :- Let $f(x) \lt 0$ ( from Case I we get $x\in (R-A)$)
Hence in the above interval we can replace $\vert f(x) \vert = -f(x) $ in the given in-equality.
We get $\vert f(x) \vert = -f(x) \le a$ Multiply both sides by $-1$
Hence we get $ f(x) \ge -a$ let interval of x is $x\in D$.
Now the interval which satisfy inequality of Case II and f(x) is negative simultaneously is $x\in (R-A)\cap D$ let this interval we call as interval E.
Hence The interval in which given inequality is satisfied is $-a\le f(x) \le a$ and the value of x will be the union of solution of case I and Case II i.e $x\in C\cup E$
Note :- If $\vert f(x) \vert \lt a$ where $ a\lt 0$ then $x\in \emptyset$
(1) Find x such that $\vert x \vert \le 2$
Sol :- If $\vert x \vert \le2$
$\Rightarrow -2\le x \le 2$
Step by step approach.
We know that $\vert x \vert = \begin{cases} x&,if\;x\ge 0 \\ -x&,if\;x\lt 0 \end{cases}$
Case I:- Let $x\ge 0$ Hence $\vert x \vert =x$ put in given in-equation we get.
$\vert x \vert = x \le 2$ that is $x\in (-\infty, 2]$
Now take intersection of $x \ge 0$ and $x\in (-\infty,2]$ we get $x\in [0,2]\;\;–(1)$ be the solution of Case I.
Case II:- Let $x\lt 0$ Hence $\vert x \vert = -x$ put in the given in-equation. We get
$\vert x \vert = -x \le 2$ multiply both sides by $-1$ we get,
$ x\ge -2$ that is $x \in [-2,\infty)$
Now take intersection of $x \lt 0$ and $x\in [-2,\infty)$ we get $x\in [-2,0)\;\;—(2)$ be the solution of Case II.
Hence final solution is Union of (1) and (2) that is $x\in [-2,2]$
(2) Find x such that $\vert \cfrac{2x}{1-x} \vert \le 1$
Sol :- Method I ( By Using Above result directly )
If $\vert \cfrac{2x}{1-x} \vert \le 1 $
$\Rightarrow -1\le \cfrac{2x}{1-x} \le 1$
From L H Inequality. we get
$\cfrac{2x}{1-x} +1\ge 0$
Hence $\cfrac{x+1}{1-x} \ge 0$ Now using wavy curve method $x\in [-1,1)\;\;—(1)$
Now Considering R H Inequality. we get
$\cfrac{2x}{1-x}\le 1\;\Rightarrow \cfrac{2x}{1-x}-1\le 0$
Hence we ge, $\cfrac{3x-1}{1-x}\le 0$ Using wavy curve method we get $x\in (-\infty, \cfrac{1}{3}]\cup (1,\infty)\;\;—(2)$
Now, taking Intersection of (1) and (2) , ( Since the variable x has to satisfy left and Right Hand Inequality simultaneously ).
Hence we get $x\in [-1,\cfrac{1}{3}]$
Method II ( By Basic approach )
We know that $\vert \cfrac{2x}{1-x} \vert =\begin{cases} \cfrac{2x}{1-x} &,\, x\in [0,1) \\ -\cfrac{2x}{1-x}&,\, x\in (-\infty,0)\cup (1,\infty) \end{cases}$
Case I :- Let $x\in [0,1)$ Now in the given interval $\vert \cfrac{2x}{1-x} \vert = \cfrac{2x}{1-x}$ put in given in-equation.
Hence we get $\vert \cfrac{2x}{1-x} \vert = \cfrac{2x}{1-x} \le 1$
$\Rightarrow \cfrac{2x}{1-x} – 1\le 0$
$\Rightarrow \cfrac{3x – 1}{1 – x}\le 0$ Now using wavy curve method we get,
$x\in (-\infty,\cfrac{1}{3}]\cup (1,\infty)$ Taking intersection of this interval with $x\in [0,1)$ we get,
$x\in [0,\cfrac{1}{3}]\;\;—(1)$
Case II :- Let $x\in (-\infty,0)\cup (1,\infty)$ then
$\vert \cfrac{2x}{1-x} \vert = -\cfrac{2x}{1-x}\le 1$ Simplifying the in-equation we get,
$\cfrac{2x}{1-x}+1\ge 0$
$\Rightarrow \cfrac{x+1}{1-x}\ge 0$ Hence we get $x\in [-1,1)$ taking intersection with $x\in (-\infty,0)\cup (1,\infty)$ we get,
$\therefore x\in [-1,0)\;\;—(2)$ is solution of Case II.
The Solution of in-equation is union of Solution of Case I and Case II i.e Union of (1) and (2). Hence Solution is
$x\in [-1,\cfrac{1}{3}]$
Example 1 :: Find solution of equation $\vert x \vert + \vert x-1 \vert = 5 $
Solution 1 :: Given $\vert x \vert + \vert x-1 \vert = 5 $
We know that $\vert x \vert = \begin{cases} x &,\, x\geq 0 \\ -x &,\, x\lt 0 \end{cases}$
Also $\vert x-1 \vert = \begin{cases} x-1 &,\, x\geq 1 \\ -(x-1) &,\, x\lt 1 \end{cases}$
Case I :: Let $x\lt 0$
Hence Given equation can be written as
$-x – (x-1)=5$ $\Rightarrow \; -2x=4$
$\therefore x=-2$ also satisfy $x\lt 0$
Case II :: let $0\leq x \lt 1$ then the given equation will be
$x – (x-1)=5$ $\Rightarrow \; 1=5$ That is not possible for any value of x.
Hence solution to this case is $x\in \emptyset$
Case III :: Let $x \geq 1$ Hence given equation will be
$x+x-1=5$ $\Rightarrow 2x=6$
Hence $x=3$ and satisfy the condition that $x\geq 1$.
Hence Solution is $x= -2 \,,\, 3$