Question

The signum function, $f$ : $R \to R$ is given by $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x=0 \\ -1, & \text{if } x < 0 \end{cases} $ is

• A. One-one
• B. Onto
• C. Many-one
• D. None of these

Answer 1

The Correct Option is C

 It can be observed that although 1 ≠ 2, f(1)=f(2)=1. F is therefore not one-one.

There is no x in domain R such that f(x)=-2 because f(x) only accepts three values for the element -2 in co-domain R: (1, 0, or -1).

F is hence not onto.

As a result, neither one-one nor onto describe the Signum function.

Signum function along with the other kinds of special functions, such as, the identity function, constant function, polynomial function, rational function and the modulus function is an important part of Mathematics. The Signum function is denoted through f(x).

Although, mathematically the signum function is defined as:

F(x) = x/|x| ---- (Equation of Signum Function)

In the above equation,

F(x) represents Signum Function which means that,

If x<0, then F(x) = -1

If x=0, then F(x) = 0

If x>0, then F(x) = 1

In other words, this means that, if the value of x is negative, the signum function will also be negative (-1), while if the value of x is positive, then, f (x) shall also be positive (1) and if the value of x is zero, the signum function shall be zero. Therefore, the value of x and f (x) are directly proportional to each other.

The graph drawn above shows a break in the curve where the value of x is zero. Whereas, apart from that break, the f(x) is seen in continuation for every value of x other than zero. Therefore, for a signum function f(x):

Domain: x ∈ R and 

Range: {-1, 0, 1}.