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

Let $$f, g :$$ $$\left[ { - 1,2} \right] \to R$$ be continuous functions which are twice differentiable on the interval $$(-1, 2)$$. Let the values of f and g at the points $$-1, 0$$ and $$2$$ be as given in the following table:

In each of the intervals $$(-1, 0)$$ and $$(0, 2)$$ the function $$(f-3g)''$$ never vanishes. Then the correct statement(s) is (are)