If $f(x)= \begin{cases}\frac{x}{1+|x|}, & |x| \geq 1 \\ \frac{x}{1-|x|}, & |x|<1\end{cases}$ then f(x) is
Answer & explanation
Correct answer: option 3
We have,
$f(x)= \begin{cases}\frac{x}{1+|x|}, & |x| \geq 1 \\ \frac{x}{1-|x|}, & |x|<1\end{cases}$
$\Rightarrow f(x)= \begin{cases}\frac{x}{1-x}, & x \leq-1 \\ \frac{x}{1+x}, & -1<x<0 \\ \frac{x}{1-x}, & 0<x<1 \\ \frac{x}{1+x}, & x \geq 1\end{cases}$
Clearly, f(x) is discontinuous and hence non-differentiable at x = ±1.
It can be easily seen that f(x) is continuous and differentiable at x = 0.