JEE MAIN - Mathematics (2006 - No. 3)
The set of points where $$f\left( x \right) = {x \over {1 + \left| x \right|}}$$ is differentiable is
$$\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)$$
$$\left( { - \infty ,1} \right) \cup \left( { - 1,\infty } \right)$$
$$\left( { - \infty ,\infty } \right)$$
$$\left( {0,\infty } \right)$$
Explanation
$$f\left( x \right) = \left\{ {\matrix{
{{x \over {1 - x}},} & {x < 0} \cr
{{x \over {1 + x}},} & {x \ge 0} \cr
} } \right.$$
$$ \Rightarrow f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right.$$
$$\therefore$$ $$f'\left( x \right)$$ exist at everywhere.
$$ \Rightarrow f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right.$$
$$\therefore$$ $$f'\left( x \right)$$ exist at everywhere.
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