Question

If \( f(x) \) is the signum function, then in terms of \( f(x) \), the constant function \( g(x)=1, \ \forall x \in \mathbb{R} \) is:

• A. \[ g(x) = \begin{cases} 2 - f(x), & x < 0 \\ f(x), & x \geq 0 \end{cases} \]
• B. \[ g(x) = \begin{cases} f(x) + f(-x), & x < 0 \\ f(x) f(-x), & x \geq 0 \end{cases} \]
• C. \[ g(x) = \begin{cases} 1 + f(x), & x>0 \\ 1 - f(x), & x \leq 0 \end{cases} \]
• D. \[ g(x) = \begin{cases} f(x) + 2, & x < 0 \\ 1 + f(x), & x = 0 \\ f(x), & x>0 \end{cases} \]

Hint:

Always recall the definition of the signum function and carefully plug in values at different intervals to verify constant behavior.

Answer 1

The Correct Option is D

We know the signum function \( f(x) \) is: \[ f(x) = \begin{cases} 1, & x>0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} \] To get a constant function \( g(x) = 1 \) for all \( x \), option (4) correctly adjusts values of \( f(x) \) at different regions of \( x \) to ensure the result is always 1: - When \( x < 0 \): \( f(x) + 2 = -1 + 2 = 1 \) - When \( x = 0 \): \( 1 + f(0) = 1 + 0 = 1 \) - When \( x>0 \): \( f(x) = 1 \)