If f(x) = begin{cases} frac{1 - [x]}{1 + x}, | vedclass

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(D) Given $f(x) = \begin{cases} \frac{1 - [x]}{1 + x}, & x 
e -1 \ 1, & x = -1 \end{cases}$.We need to evaluate $f(|2k|)$. Let $y = |2k|$. Since $|2

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(D) Given $f(x) = \begin{cases} \frac{1 - [x]}{1 + x}, & x 
e -1 \ 1, & x = -1 \end{cases}$.We need to evaluate $f(|2k|)$. Let $y = |2k|$. Since $|2k| \ge 0$,we consider $x \ge 0$ for $f(x) = \frac{1 - [x]}{1 + x}$.For $0 \le x At $x = 0$,$f(0) = 1$. The limit $\lim_{x 	o 0} f(x) = 1$,so it is continuous at $x = 0$.At $x = -1$,the function is defined as $1$. Checking the limit $\lim_{x 	o -1} \frac{1 - [x]}{1 + x}$,as $x 	o -1^+$,$[x] = -1$,so $\frac{1 - (-1)}{1 + x} = \frac{2}{1 + x} 	o \infty$. Thus,it is discontinuous at $x = -1$.For $x = \frac{1}{2}$,$f(x) = \frac{1 - [1/2]}{1 + 1/2} = \frac{1 - 0}{1.5} = \frac{2}{3}$. As $x 	o \frac{1}{2}^-$,$f(x) 	o \frac{2}{3}$. As $x 	o \frac{1}{2}^+$,$[x] = 0$,$f(x) 	o \frac{2}{3}$. However,if we consider the function $f(|2k|)$,the jump occurs at values where $[2k]$ changes,i.e.,$2k = n$. Thus,$k = n/2$. At $k = 1/2$,$2k = 1$,$[2k]$ jumps from $0$ to $1$,causing discontinuity.Therefore,all statements provided in the options describe properties of the function or its behavior at specific points.

If $f(x) = \begin{cases} \frac{1 - [x]}{1 + x}, & x \ne -1 \\ 1, & x = -1 \end{cases}$,then the value of $f(|2k|)$ will be (where $[.]$ denotes the greatest integer function). Which of the following statements is true?

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