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Modulus. - Best Maths Practice Material

Modulus.

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 $x= -2 \,,\, 3$