Examples on Modulus Function

What is the Modulus Function?

The modulus of a given number describes the magnitude of the number. Modulus Function is defined as the real valued function, say f : R -> R, where y = |x| for each x ∈ R OR f(x) = |x|. This function can be defined using modulus operation as follows:

f(x)={xx<0xx0f(x) =\left\{\begin{matrix} -x &x <0 \\ x & x \geq 0 \end{matrix}\right.

For each non-negative value of x, f(x) is equal to the positive value of x. But for negative values of x, f(x) is equal to negative value of x.

Properties of Modulus Function

The modulus function has the following properties

1. For any real number x, we have
x2=x\sqrt{x^2} = |x|

2. x=x||x||=|x|

3. If a and b are positive real numbers, then
a] x2a2xaaxab] x2a2xaxac] x2>a2x>ax<ad] a2x2b2axbxϵ[b,a][a,b].a]\ x^2 \leq a^2 \Leftrightarrow |x| \leq a \Leftrightarrow -a \leq x \leq a\\ b] \ x^2 \geq a^2 \Leftrightarrow |x| \geq a \Leftrightarrow x \leq -a \\ c] \ x^2 > a^2 \Leftrightarrow |x| > a \Leftrightarrow x < -a \\ d]\ a^2 \leq x^2 \leq b^2 \Leftrightarrow a \leq |x| \leq b \Leftrightarrow x \epsilon [−b, −a] \cup [a, b].\\

4.  If a is negative, then
a. xa|x| \neq a, x in Rb.|x| ≤ a, x = ϕ

5.  For real number x and y
a.xy=xyb.xy=xy,y0a. |xy| = |x| |y|\\ b. |xy| = |x| |y|, y \neq 0|\\ c.x+yx+y and xyxy for x, y0 or x, y<0,x+y=x+yc. |x+y| \leq |x| + |y| \text \ and \ |x−y| \leq |x| – |y| \text \ for \ x, \ y \geq 0 \text \ or \ x, \ y < 0, |x+y| = |x| + |y|\\ d.x+yxyx+y andxyxy for x, y0 and xyy or x, y<0 and xy,xy=xyd. |x+y| \geq |x| − |y| |x+y| \text \ and |x−y| \geq |x| − |y| \text \ for \ x, \ y \geq 0 \text \ and \ |x| \geq |y| y \text \ or \ x, \ y < 0 \text \ and \ |x| \geq |y|, |x−y| = |x| − |y|\\

6. The absolute value has the following four fundamental properties (ab are real numbers),

1] a0 Nonnegativity2] a=0    a=0  Positivedefiniteness3] ab=ab Multiplicativity4] a+ba+b Subadditivity,specifically the triangle inequality1] \ {\displaystyle |a|\geq 0} \text \ Non-negativity\\ 2] \ {\displaystyle |a|=0\iff a=0} \text  \ Positive-definiteness\\ 3] \ {\displaystyle |ab|=|a|\,|b|} \text \ Multiplicativity\\ 4] \ {\displaystyle |a+b|\leq |a|+|b|} \text \ Subadditivity, \text specifically \ the \ triangle \ inequality\\

7. Some additional useful properties are given below. These are either immediate consequences of the definition or implied by the four fundamental properties above.

a] a=a Idempotence (the absolute value of the absolute value is the absolute value)b] a=a Evenness (reflection symmetry of the graph)c] ab=0    a=b Identity of indiscernibles (equivalent to positivedefiniteness)d] abac+cb Triangle inequality (equivalent to subadditivity)e] ab=ab  ( if b0) Preservation of division (equivalent to multiplicativity)f] abab Reverse triangle inequality (equivalent to subadditivity) a]\ {\displaystyle {\big |}\,|a|\,{\big |}=|a|} \text \ Idempotence \ (the \ absolute \ value \ of \ the \ absolute \ value \ is \ the \ absolute \ value)\\ b] \ {\displaystyle |-a|=|a|} \text \ Evenness \ (reflection \ symmetry \ of \ the \ graph)\\ c] \ {\displaystyle |a-b|=0\iff a=b} \text \ Identity \ of \ indiscernibles \ (equivalent \ to \ positive-definiteness) \\ d] \ {\displaystyle |a-b|\leq |a-c|+|c-b|} \text \ Triangle \ inequality \ (equivalent \ to \ subadditivity) \\ e] \ {\displaystyle \left|{\frac {a}{b}}\right|={\frac {|a|}{|b|}}\ }\ (\text\ if \ {\displaystyle b\neq 0}) \text \ Preservation \ of \ division \ (equivalent \ to \ multiplicativity) \\ f] \ {\displaystyle |a-b|\geq {\big |}\,|a|-|b|\,{\big |}} \text\ Reverse \ triangle \ inequality \ (equivalent \ to \ subadditivity)\\

8. Two other useful properties concerning inequalities are:

1] ab    bab2] ab    ab  or ab1] \ {\displaystyle |a|\leq b\iff -b\leq a\leq b}\\ 2] \ {\displaystyle |a|\geq b\iff a\leq -b\ } \text \ or \ {\displaystyle a\geq b}\\

Graph of Modulus Function

Modulus Graph Function

Solved Examples on Modulus

Example 1: Solve modulus and find the interval of x for |x2 – 5x + 6|

Solution: |x2 – 5x + 6| = |(x – 2)(x – 3)| = |f(x)|

As per modulus definition,

|f(x)| = f(x); if f(x) is positive

| f(x) |= -f(x); if f(x) is negative

f(x) = (x – 2)(x – 3) is positive or zero when x = (- ∞, 2] ∪ [3, ∞)

f(x) = (x – 2)(x – 3) is negative when x = (2, 3)

So, |x2 – 5x + 6| = (x2 – 5x + 6) when x = (-∞, 2] ∪ [3, ∞) and

|x2 – 5x + 6| = -(x2 – 5x + 6) when x = (2, 3)

Example 2: If |x2 – 5x + 6| + |x2 – 8x + 12| = 0. Find x.

Solution:

Every modulus is a non-negative number and if two non-negative numbers add up to get zero then individual numbers itself equal to zero simultaneously.

x2 – 5x + 6 = 0 for x = 2 or 3

x2 – 8x + 12 = 0 for x = 2 or 6

Both the equations are zero at x = 2

So, x = 2 is the only solution for this equation.

Example 3: Solve for x, |x – 1| – |x – 2| = 10

Solution: Here the critical points are 1 and 2.

Let us check for the values less than 1, between 1 and 2, and greater than 2.

Case 1: For x ≤ 1

-(x – 1) – {-(x – 2)} = 10

or -x + 1 + x – 2 = 10

or -1 = 10 (which is not possible)

Case 2: x ∈ (1, 2)

(x – 1) – {-(x – 2)} = 10

or x – 1 + x – 2 = 10

or 2x – 3 = 10

or x = 13/2

In this case x ∈ (1, 2), so 6.5 is not a solution.

Case 3: x ≻ 2

(x – 1) – (x – 2) = 10

=> 1 = 10 (which is not possible)

So this equation has no solution.

Example 4: Draw the graph of |sin(x) + cos(x)| in the interval x ϵ [0, π].

Solution:

|sin(x) + cos(x)| = |√2 sin(x + π/4)|

First, make the graph of the function then take the modulus

Here at the place of sinx we have √2 sin(x + π/4) so, the graph will be of type sin x only but the graph will be starting from x = -π/4.

y = sin(x) + cos(x)

Modulus JEE

Take all negative y value to the positive y side and positive y remains the same

y = |sin(x) + cos(x)| = |√2 sin(x + π/4)|

Example on Modulus

Example 5: Find the domain and range of the below function?
a. y = |1−x|
b. y = 2 − |1 − x| y
c. y = 2√x − |x|

Solution:

a. y = |1 − x|
Clearly, this is defined for x ∈ R. So Domain is R.
Now |1 − x| ≥ 0 for all x ∈ R
So Range is [0, ∞)

b. y = 2 − |1 − x|
Clearly, this is defined for x ∈ R. So Domain is R.
Now |1 − x| ≥ 0 for all x ∈ R
or
−|1 − x| ≤ 0
2 − |1 − x| ≤ 2 for all x∈R
So Range is (−∞, 2]

c. y = 2√x − |x|
Now x− |x| = {x − x = 0 if x ≥ 0x + x = 2x if x ≤ 0

Therefore, 2√x − |x| is undefined for all x ∈ R
So the domain of the function is ϕ.