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1. The range of the function [tex]$f(x)=\frac{H(x)}{2}$[/tex], where [tex]$H(x)$[/tex] is the Heaviside function, is:

A. [tex]$(0, \infty)$[/tex]

B. [tex]$(-\infty, \infty)$[/tex]

C. [tex]$\{2\}$[/tex]

D. [tex]$\left\{0, \frac{1}{2}\right\}$[/tex]


Sagot :

To determine the range of the function [tex]\( f(x) = \frac{H(x)}{2} \)[/tex] where [tex]\( H(x) \)[/tex] is the Heaviside function, let's start by understanding the Heaviside function itself.

The Heaviside function [tex]\( H(x) \)[/tex] is defined as follows:
[tex]\[ H(x) = \begin{cases} 0 & \text{if } x < 0 \\ 1 & \text{if } x \geq 0 \end{cases} \][/tex]

Now, let's analyze the given function [tex]\( f(x) = \frac{H(x)}{2} \)[/tex]:
[tex]\[ f(x) = \frac{H(x)}{2} = \begin{cases} \frac{0}{2} = 0 & \text{if } x < 0 \\ \frac{1}{2} = \frac{1}{2} & \text{if } x \geq 0 \end{cases} \][/tex]

From the above definition, we can see that:
- When [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 0 \)[/tex].
- When [tex]\( x \geq 0 \)[/tex], [tex]\( f(x) = \frac{1}{2} \)[/tex].

Thus, the function [tex]\( f(x) = \frac{H(x)}{2} \)[/tex] can only take on two different values depending on whether [tex]\( x \)[/tex] is negative or non-negative:
[tex]\[ f(x) \in \{0, \frac{1}{2}\} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{\left\{0, \frac{1}{2}\right\}} \][/tex]

This corresponds to option (D). So, the range of the function [tex]\( f(x) = \frac{H(x)}{2} \)[/tex] is [tex]\(\left\{ 0, \frac{1}{2} \right\} \)[/tex].