# Modulus Function

Function y = f(x) = |x| is known as modulus function.

y = f(x) = $\left\{\begin{matrix}x,&x\geqslant 0\\-x,& x<0 \end{matrix} \right.$

Domain of f(x) = x $\in$ R

and Range of f(x) = [0, )

Properties of Modulus Function

Since the modulus function can be effective to find inequality between the numbers, here are the following properties of the modulus function:

Here are some other non-negative expressions that can explain the non-negative value of the modulus function:

1. The even exponent of an expression or variable can be defined as x2n , where n ∈ Z.

2. The even root of a variable can be defined as x1/2n, where n ∈ Z.

3. The value of y can be defined as y = 1−sinx or y = 1−cosx, (since sinx ≤ 1 and cosx ≤1)

• When a > 0,

Here, x lies between −a and a, not considering the endpoints of the interval, i.e.,

|x| < a; a > 0 ⇒ −a < x < a

|x| > a; a > 0  ⇒ x < − a or x > a ⇒ x ∈ (−∞, −a) ∪ (a, ∞)

Since the inequalities can be useful to express intervals in the compact form, here's an example of the cosec trigonometric function range that is defined as x ∈ (−∞, −1] ∪ [1,∞}, represented as:

|x| ≥ 1

|f(x)| < a; a > 0, ⇒ −a < f(x) < a

• For x, y as real variables:

1. |x − y| = 0, ⇔ x = y

2. |x + y| ≤ |x| + |y|

3. |x − y| ≥ ||x| − |y||

4. |xy| = |x| * |y|

5. |x/y| = |x| / |y|, where |y| ≠ 0.

• For a and b as positive real numbers:

1. x2 ≤ a2 ⇔ |x| ≤ a ⇔ −a ≤ x ≤ a

2. x2 ≥ a2 ⇔ |x| ≥ a ⇔ x ≤ −a , x ≥ a

3. x2 < a2 ⇔ |x| < a ⇔ −a < x < a

4. x2 > a2 ⇔ |x| > a ⇔ x < −a, x > a

5. a2 ≤ x2 ≤ b2 ⇔ a ≤ |x| ≤ b ⇔ x ∈ −b,aa,b, . . .

6. a2 < x2 <b2⇔ a < |x| < b ⇔ x ∈ (−b, −a) ∪ (a, b)

Sample Problem 1:

Solve |2x + 1| = 7 using modulus function.

We know that the modulus function always gives a non-negative output, therefore we have two cases:

If 2x + 1 > 0, then |2x + 1| = 2x + 1 and

If 2x + 1 < 0, then |2x + 1| = −(2x + 1)

Case 1: If 2x + 1 > 0, we have

|2x + 1| = 2x + 1

⇒ 2x + 1 = 7

⇒ 2x = 7 − 1 = 6

⇒ x = 3

Case 2: If 2x + 1 < 0, we have

|2x + 1| = − (2x + 1)

⇒ − (2x + 1) = 7

⇒ 2x + 1 = −7

⇒ 2x = −7 − 1 = −8

⇒ x = −4

Hence, the solution for x is −4 and 3.

We can say −4 < x < 3.

Sample Problem 2:

If |x2 – 7x + 10| + |x2 – 6x + 8| = 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 – 7x + 10 = 0 for x = 2 or 5

x2 – 6x + 8 = 0 for x = 2 or 4

Both the equations are zero at x = 2

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

Sample Problem 3:

Solve for x, |x – 1| – |x – 2| = 10

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 13/2 is not a solution.

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