Let $${f_1}:R \to R,\,{f_2}:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R,\,{f_3}:( - 1,{e^{\pi /2}} - 2) \to R$$ and $${f_4}:R \to R$$ be functions defined by

(i) $${f_1}(x) = \sin (\sqrt {1 - {e^{ - {x^2}}}} )$$,

(ii) $${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {\tan { - ^1}x}}if\,x \ne 0,\,where \hfill \cr 1\,if\,x = 0 \hfill \cr} \right.$$

the inverse trigonometric function tan^$$-$$1x assumes values in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$,

(iii) $${f_3}(x) = [\sin ({\log _e}(x + 2))]$$, where for $$t \in R,\,[t]$$ denotes the greatest integer less than or equal to t,

(iv) $${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right)\,if\,x \ne 0 \hfill \cr 0\,if\,x = 0 \hfill \cr} \right.$$

LIST-I	LIST-II
P. The function $$ f_1 $$ is	1. NOT continuous at $$ x = 0 $$
Q. The function $$ f_2 $$ is	2. continuous at $$ x = 0 $$ and NOT differentiable at $$ x = 0 $$
R. The function $$ f_3 $$ is	3. differentiable at $$ x = 0 $$ and its derivative is NOT continuous at $$ x = 0 $$
S. The function $$ f_4 $$ is	4. differentiable at $$ x = 0 $$ and its derivative is continuous at $$ x = 0 $$

If f : R $$ \to $$ R is a twice differentiable function such that f"(x) > 0 for all x$$ \in $$R, and $$f\left( {{1 \over 2}} \right) = {1 \over 2}$$, f(1) = 1, then

Let $$f(x) = \left\{ {\matrix{ {{x^2}\left| {\cos {\pi \over x}} \right|,} & {x \ne 0} \cr {0,} & {x = 0} \cr } } \right.$$

x$$\in$$R, then f is