Let [x] denote the greatest integer less than or equal to x. Then $$\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$

Let $$f\left( x \right) = \left\{ {\matrix{ { - 1} & { - 2 \le x < 0} \cr {{x^2} - 1,} & {0 \le x \le 2} \cr } } \right.$$ and

$$g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).$$

Then, in the interval (–2, 2), g is :

Let f : ($$-$$1, 1) $$ \to $$ R be a function defined by f(x) = max $$\left\{ { - \left| x \right|, - \sqrt {1 - {x^2}} } \right\}.$$ If K be the set of all points at which f is not differentiable, then K has exactly -

For each t $$ \in $$ R , let [t] be the greatest integer less than or equal to t

Then  $$\mathop {\lim }\limits_{x \to 1^ + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}$$