Let f(x) = begin{cases} x sin left(frac{1}{x} | vedclass

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(A) Given the function $f(x) = \begin{cases} x \sin \left(\frac{1}{x}ight) & 	ext{when } x 
eq 0 \ 1 & 	ext{when } x = 0 \end{cases}$.We want to

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exactly one element

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(A) Given the function $f(x) = \begin{cases} x \sin \left(\frac{1}{x}ight) & 	ext{when } x 
eq 0 \ 1 & 	ext{when } x = 0 \end{cases}$.We want to find the set $A = \{x \in \mathbb{R} : f(x) = 1\}$.Case $1$: If $x = 0$,then $f(0) = 1$. Thus,$0 \in A$.Case $2$: If $x 
eq 0$,we solve $x \sin \left(\frac{1}{x}ight) = 1$,which implies $\sin \left(\frac{1}{x}ight) = \frac{1}{x}$.Let $t = \frac{1}{x}$. Then the equation becomes $\sin(t) = t$.We know that for all $t 
eq 0$,$|\sin(t)| Specifically,if $t > 0$,$\sin(t)  t$.Therefore,the equation $\sin(t) = t$ has only one solution at $t = 0$.However,our substitution was $t = \frac{1}{x}$,and $x 
eq 0$ implies $t$ cannot be $0$.Thus,there are no solutions for $x 
eq 0$.Consequently,the only element in set $A$ is ${0}$.Therefore,$A$ has exactly one element.

Let $f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right) & \text{when } x \neq 0 \\ 1 & \text{when } x = 0 \end{cases}$ and $A = \{x \in \mathbb{R} : f(x) = 1\}$. Then,$A$ has

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