A) 0
B) 1
C) 2
D) does not exist
Correct Answer: B
Solution :
Let \[x<0\Rightarrow |x|=-x\] \[\Rightarrow \]\[f(x)=\frac{d}{dx}\left( \frac{x}{1-x} \right)=\frac{1}{{{(1-x)}^{2}}}\] \[\Rightarrow \] \[{{[f(x)]}_{x=0}}=1\] Again \[x>0\Rightarrow |x|=x\] \[f(x)=\frac{d}{dx}\left( \frac{x}{1+x} \right)\] \[=\frac{1}{{{(1+x)}^{2}}}\] \[\Rightarrow \] \[{{[f(x)]}_{x=0}}=1\] \[\Rightarrow \] \[f(0)=1\]You need to login to perform this action.
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