If ⨍ is defined on D=R~-1,1 by ⨍ x=x1-x then

fx=x1-x ∀x∈R--1, 1 =-x1+x ;x<0x1-x ; x≥0 f10-=limx→0--x1+x-0x-0=-1 f10+=limx→0+x1+x-0x-0=1 ∴f10≠f10+ ∴f is differentiable in D-0

If ⨍ is defined on D=R~-1,1 by ⨍ x=x1-x then

fx=x1-x ∀x∈R--1, 1 =-x1+x ;x<0x1-x ; x≥0 f10-=limx→0--x1+x-0x-0=-1 f10+=limx→0+x1+x-0x-0=1 ∴f10≠f10+ ∴f is differentiable in D-0

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If $⨍$ is defined on $D = R ~ \{- 1, 1\}$ by $⨍ (x) = \frac{|x|}{1 - |x|}$ then

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$⨍$ is continuous on $D$

$⨍$ is continuous but not differentiable on $D$

$⨍$ is differentiable at $D ~ \{0\}$

$⨍$ is differentiable but not continuous on $R$

answer is C.

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Detailed Solution

$f (x) = \frac{|x|}{1 - |x|} \forall x \in R - \{- 1, 1\} = \{\begin{cases} \frac{- x}{1 + x}; x < 0 \\ \frac{x}{1 - x}; x \geq 0 \end{cases} f^{1} (0^{-}) = \lim_{x \to 0^{-}} \frac{\frac{- x}{1 + x} - 0}{x - 0} = - 1 f^{1} (0^{+}) = \lim_{x \to 0^{+}} \frac{\frac{x}{1 + x} - 0}{x - 0} = 1 ∴ f^{1} (0) \neq f^{1} (0^{+}) ∴ f is differentiable in D - \{0\}$