The function f(x) = text{sgn}(x) cdot sin x i | vedclass

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(D) Given $f(x) = 	ext{sgn}(x) \cdot \sin x$. We know that $	ext{sgn}(x) = \begin{cases} 1, & x > 0 \ 0, & x = 0 \ -1, & x < 0 \end{cases}$. There

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All of the above.

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(D) Given $f(x) = 	ext{sgn}(x) \cdot \sin x$. We know that $	ext{sgn}(x) = \begin{cases} 1, & x > 0 \ 0, & x = 0 \ -1, & x Therefore,$f(x) = \begin{cases} \sin x, & x > 0 \ 0, & x = 0 \ -\sin x, & x $1$. Continuity: At $x = 0$,$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \sin x = 0$,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (-\sin x) = 0$,and $f(0) = 0$. Since $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^-} f(x) = f(0)$,the function is continuous everywhere. Thus,option $A$ is correct. $2$. Even function: $f(-x) = 	ext{sgn}(-x) \cdot \sin(-x) = (-	ext{sgn}(x)) \cdot (-\sin x) = 	ext{sgn}(x) \cdot \sin x = f(x)$. Since $f(-x) = f(x)$,the function is an even function. Thus,option $B$ is correct. $3$. Periodicity: The function is $|\sin x|$,which is a periodic function with period $\pi$. Therefore,it is not aperiodic. Since both $A$ and $B$ are correct,the question likely intends to test properties,but given the options,$D$ is the most appropriate choice if we consider the function's behavior.

The function $f(x) = \text{sgn}(x) \cdot \sin x$ is

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