Let, $$f(x) = {{\sin \pi x} \over {{x^2}}}$$, x > 0

Let x₁ < x₂ < x₃ < ... < x_n < ... be all the points of local maximum of f and y₁ < y₂ < y₃ < ... < y_n < ... be all the points of local minimum of f.

Then which of the following options is/are correct?

f : R $$ \to $$ R is a differentiable function such that f'(x) > 2f(x) for all x$$ \in $$R, and f(0) = 1 then

If $$f(x) = \left| {\matrix{ {\cos 2x} & {\cos 2x} & {\sin 2x} \cr { - \cos x} & {\cos x} & { - \sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right|$$,

then

Let f: R $$ \to \left( {0,\infty } \right)$$ and g : R $$ \to $$ R be twice differentiable functions such that f'' and g'' are continuous functions on R. Suppose f'$$(2)$$ $$=$$ g$$(2)=0$$, f''$$(2)$$$$ \ne 0$$ and g'$$(2)$$ $$ \ne 0$$. If
$$\mathop {\lim }\limits_{x \to 2} {{f\left( x \right)g\left( x \right)} \over {f'\left( x \right)g'\left( x \right)}} = 1,$$ then