If f(x) = cases 1-|x| 1+x , & x -1 1, & x = -1 cases then the value of will be (where…

MATHEMATICS · Single correct · AoPS · Mathematics · Special Functions · Special Functions

Problem

If f(x)={1−∣x∣1+x,x≠−11,x=−1f(x) = \begin{cases} \frac{1-|x|}{1+x}, & x \neq -1 \\ 1, & x = -1 \end{cases} then the value of f([2x])f([2x]) will be (where [⋅][\cdot] shows the greatest integer function) (a) Continuous at x=−1x = -1 (b) Continuous at x=0x = 0 (c) Discontinuous at x=12x = \frac{1}{2} (d) All of these

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