$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}$$ is equal to

Consider the function, $$f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|,x \in R$$

Statement - 1 : $$f'\left( 4 \right) = 0$$

Statement - 2 : $$f$$ is continuous in [2, 5], differentiable in (2, 5) and $$f$$(2) = $$f$$(5)

If $$f:R \to R$$ is a function defined by

$$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$,

where [x] denotes the greatest integer function, then $$f$$ is

$$\mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)$$